Math

烙饼 tutorial 设 $a_i$ 的和是 $sum$，每个机器的平均使用时间是 $avg = \lceil \frac{sum}{m} \rceil$，如果 $mx = max(a_i) > avg$，此时最优策略是让 $mx$ 一直在一台机器上，因为移动它不会让结果更好，此时的总时间就是 $mx$。如果 $mx \le avg$，此时总时间是 $avg$，从第一个机器开始考虑，因为 $a_i \le mx \le avg$，因此可以把 $a_i$ 放在第一个机器上，如果第一个机器还有剩余时间，把剩余的时间给 $a_{i + 1}$，然后对第二个机器，从时间 $0$ 开始，把 $a_{i + 1}$ 剩余的时间用完，此时可以保证 $a_{i + 1}$ 在两个机器上的时间是不会重合的，如果重合，这意味着 $a_{i + 1}$ 占用了前一个机器的后半段时间和后一个机器的前半段时间，并且这两段时间的和大于 $avg$，这和假设矛盾。因此每个 $a_i$ 最多产生两个记录。 接下来考虑为什么是取上整：$avg = \lceil \frac{sum}{m} \rceil$。如果 $sum \% m \ne 0$，此时意味着存在一个 $a_i$ 还有剩余时间 $r_i, 1 \le r_i \lt m$，设 $avg_1 = \lfloor \frac{sum}{m} \rfloor$。一种方案是在最后一个机器上一直等 $a_i$ 完成，此时总时间是 $avg_1 + r_i$。另一种方案是把 $r_i$ 均摊到 $m$ 个机器上，也就是每个机器多用 $1$ 单位的时间，实际时间是 $avg_1 + 1$，并且总的可用时间多了 $m$，能够保证 $(avg_1 + 1) \cdot m \gt sum$，也就是一定可以完成所有的 $a_i$。 ...

Balance (Easy version) tutorial 对于询问的每一个 $k$，记录它的 $MEX$ 值，第二次询问的时候，直接从上次记录的值开始找第一个不在集合中的数，然后更新记录的 $MEX(k)$ 值。插入操作也是类似的，插入 $x$ 的时候，可以顺便记录 $MEX(x)$ 的值。 // Date: Fri Jan 5 13:01:33 2024 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; template <class T> using pq = priority_queue<T>; template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7, MOD1 = 998'244'353; const ll INFL = 0x3f3f3f3f'3f3f3f3f; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nemp(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) #define NL cout << '\n'; template <class T> bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; } template <class T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; } template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } void __print(int x) { cerr << x; } void __print(long x) { cerr << x; } void __print(long long x) { cerr << x; } void __print(unsigned x) { cerr << x; } void __print(unsigned long x) { cerr << x; } void __print(unsigned long long x) { cerr << x; } void __print(float x) { cerr << x; } void __print(double x) { cerr << x; } void __print(long double x) { cerr << x; } void __print(char x) { cerr << '\'' << x << '\''; } void __print(const char *x) { cerr << '\"' << x << '\"'; } void __print(const string &x) { cerr << '\"' << x << '\"'; } void __print(bool x) { cerr << (x ? "true" : "false"); } template <typename T, typename V> void __print(const pair<T, V> &x) { cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}'; } template <typename T> void __print(const T &x) { int f = 0; cerr << '{'; for (auto &i : x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}"; } void _print() { cerr << "]\n"; } template <typename T, typename... V> void _print(T t, V... v) { __print(t); if (sizeof...(v)) cerr << ", "; _print(v...); } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #define dbg(x...) \ cerr << "\e[91m" << __func__ << ":" << __LINE__ << " [" << #x << "] = ["; \ _print(x); \ cerr << "\e[39m" << endl; #else #define debug1 #define debug2 #define debug3 #define dbg(x...) #endif void solve() { int q; cin >> q; map<ll, ll> m; set<ll> s; while (q--) { ll x, k; string op; cin >> op; if (op == "+") { cin >> x; s.insert(x); ll cur = x; while (has(s, cur)) { cur += x; if (cur < x) break; } m[x] = cur; } else { cin >> k; ll cur = k; if (has(m, k)) { cur = m[k]; } while (has(s, cur)) { cur += k; } cout << cur << '\n'; m[k] = cur; } } } int main(void) { #ifdef _DEBUG freopen("1732d1.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int T = 1; while (T--) { solve(); } return 0; }

Divisible Numbers (easy version) tutorial 需要找到 $ab \mid xy$，根据整除的性质：如果 $(p, m) = 1, p \mid nm$，那么 $p \mid n$。由于 $(\frac{ab}{(ab, x)}, x) = 1$，因此 $\frac{ab}{(ab, x)} | y$。也就是需要找到 $k$，使得 $\frac{ab}{(ab, x)} k= y$，可以枚举 $x$，然后二分答案 $k$。 // Date: Thu Jan 4 22:39:23 2024 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; template <class T> using pq = priority_queue<T>; template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7, MOD1 = 998'244'353; const ll INFL = 0x3f3f3f3f'3f3f3f3f; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nemp(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) #define NL cout << '\n'; template <class T> bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; } template <class T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; } template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } void __print(int x) { cerr << x; } void __print(long x) { cerr << x; } void __print(long long x) { cerr << x; } void __print(unsigned x) { cerr << x; } void __print(unsigned long x) { cerr << x; } void __print(unsigned long long x) { cerr << x; } void __print(float x) { cerr << x; } void __print(double x) { cerr << x; } void __print(long double x) { cerr << x; } void __print(char x) { cerr << '\'' << x << '\''; } void __print(const char *x) { cerr << '\"' << x << '\"'; } void __print(const string &x) { cerr << '\"' << x << '\"'; } void __print(bool x) { cerr << (x ? "true" : "false"); } template <typename T, typename V> void __print(const pair<T, V> &x) { cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}'; } template <typename T> void __print(const T &x) { int f = 0; cerr << '{'; for (auto &i : x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}"; } void _print() { cerr << "]\n"; } template <typename T, typename... V> void _print(T t, V... v) { __print(t); if (sizeof...(v)) cerr << ", "; _print(v...); } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #define dbg(x...) \ cerr << "\e[91m" << __func__ << ":" << __LINE__ << " [" << #x << "] = ["; \ _print(x); \ cerr << "\e[39m" << endl; #else #define debug1 #define debug2 #define debug3 #define dbg(x...) #endif ll a, b, c, d; ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); } void solve() { cin >> a >> b >> c >> d; For1(x, a + 1, c) { ll g = gcd(a * b, x), g1 = a * b / g; ll lo = 1, hi = 1e5 + 10, mid; while (lo < hi) { mid = (lo + hi) / 2; if (mid * g1 > b) hi = mid; else lo = mid + 1; } if (hi * g1 > b && hi * g1 <= d) { cout << x << ' ' << hi * g1 << '\n'; return; } } cout << "-1 -1\n"; } int main(void) { #ifdef _DEBUG freopen("1744e1.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int T = 1; cin >> T; while (T--) { solve(); } return 0; }

Living Sequence tutorial 如果只能用 $0, 1$，要求第 $n$ 个数字，那么答案就是 $n$ 的二进制表示。要求不能用 $4$ 这个数字，可以把 $5, 6, 7, 8, 9$ 映射成 $4, 5, 6, 7, 8$，这样就是 9 进制数字，并且是连续的，第 $n$ 个数字就是 $n$ 的 9 进制表示，然后再把数码映射回去。 // Date: Thu Jan 4 22:12:10 2024 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; template <class T> using pq = priority_queue<T>; template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7, MOD1 = 998'244'353; const ll INFL = 0x3f3f3f3f'3f3f3f3f; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nemp(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) #define NL cout << '\n'; template <class T> bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; } template <class T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; } template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } void __print(int x) { cerr << x; } void __print(long x) { cerr << x; } void __print(long long x) { cerr << x; } void __print(unsigned x) { cerr << x; } void __print(unsigned long x) { cerr << x; } void __print(unsigned long long x) { cerr << x; } void __print(float x) { cerr << x; } void __print(double x) { cerr << x; } void __print(long double x) { cerr << x; } void __print(char x) { cerr << '\'' << x << '\''; } void __print(const char *x) { cerr << '\"' << x << '\"'; } void __print(const string &x) { cerr << '\"' << x << '\"'; } void __print(bool x) { cerr << (x ? "true" : "false"); } template <typename T, typename V> void __print(const pair<T, V> &x) { cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}'; } template <typename T> void __print(const T &x) { int f = 0; cerr << '{'; for (auto &i : x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}"; } void _print() { cerr << "]\n"; } template <typename T, typename... V> void _print(T t, V... v) { __print(t); if (sizeof...(v)) cerr << ", "; _print(v...); } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #define dbg(x...) \ cerr << "\e[91m" << __func__ << ":" << __LINE__ << " [" << #x << "] = ["; \ _print(x); \ cerr << "\e[39m" << endl; #else #define debug1 #define debug2 #define debug3 #define dbg(x...) #endif void solve() { ll n; cin >> n; char a[10] = {0, 1, 2, 3, 5, 6, 7, 8, 9}; string s; while (n) { s += '0' + (n % 9); n /= 9; } reverse(all(s)); for (auto &c : s) { int idx = c - '0'; c = '0' + a[idx]; } cout << s << '\n'; } int main(void) { #ifdef _DEBUG freopen("1811e.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int T = 1; cin >> T; while (T--) { solve(); } return 0; }

Almost All Multiples tutorial 因为 $x$ 放在了第一个位置，所以第 $x$ 的位置只能放数字 $kx$，其中 $k \ge 2$，那么第 $kx$ 个位置只能放 $k_1x$，其中 $k_1 \ge k + 1$，最终某个位置 $k_j$ 会放 $n = k_j x$，为了让字典序最小，$n$ 应该尽量放在后面，也就是让这个序列尽量长，所以把 $n/x$ 质因数分解，每次只增加一个质因子：$k_{i+1} = k_{i} p_i$，就可以让这个 $k_j$ 序列最长了。 // Date: Wed Jan 3 23:57:46 2024 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; template <class T> using pq = priority_queue<T>; template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7; const ll INFL = 0x3f3f3f3f'3f3f3f3f; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nemp(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) #define NL cout << '\n'; template <class T> bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; } template <class T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; } template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } void __print(int x) { cerr << x; } void __print(long x) { cerr << x; } void __print(long long x) { cerr << x; } void __print(unsigned x) { cerr << x; } void __print(unsigned long x) { cerr << x; } void __print(unsigned long long x) { cerr << x; } void __print(float x) { cerr << x; } void __print(double x) { cerr << x; } void __print(long double x) { cerr << x; } void __print(char x) { cerr << '\'' << x << '\''; } void __print(const char *x) { cerr << '\"' << x << '\"'; } void __print(const string &x) { cerr << '\"' << x << '\"'; } void __print(bool x) { cerr << (x ? "true" : "false"); } template <typename T, typename V> void __print(const pair<T, V> &x) { cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}'; } template <typename T> void __print(const T &x) { int f = 0; cerr << '{'; for (auto &i : x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}"; } void _print() { cerr << "]\n"; } template <typename T, typename... V> void _print(T t, V... v) { __print(t); if (sizeof...(v)) cerr << ", "; _print(v...); } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #define dbg(x...) \ cerr << "\e[91m" << __func__ << ":" << __LINE__ << " [" << #x << "] = ["; \ _print(x); \ cerr << "\e[39m" << endl; #else #define debug1 #define debug2 #define debug3 #define dbg(x...) #endif const int N = 200100; int n, a[N], x; VI defact(int n) { VI res; for (ll i = 2; i * i <= n; ++i) { while (n % i == 0) { res.pb(i); n /= i; } } if (n) res.pb(n); sort(all(res)); return res; } void solve() { cin >> n >> x; if (n % x) { cout << "-1\n"; return; } For1(i, 1, n) { if (i == 1) a[i] = x; else if (i == n) a[i] = 1; else a[i] = i; } VI ve = defact(n / x); deque<int> de(all(ve)); int cur = x; while (cur != n) { int t = de.front(); de.pop_front(); a[cur] = cur * t; cur = a[cur]; } For1(i, 1, n) { cout << a[i] << ' '; } NL; } int main(void) { #ifdef _DEBUG freopen("1758c.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int T = 1; cin >> T; while (T--) { solve(); } return 0; }

Pull Your Luck tutorial 对于 $p$，走的步数长度是 $\frac{p (p + 1)}{2}$，最终的位置是 $pos = x + \frac{p(p + 1)}{2} \mod n$，要找出一个 $p$，使得 $pos = 0$，当 $p = 2n$ 时，$pos = x + n(2n+1) \mod n$，此时 $pos, x$ 对模 $n$ 同余，也就是 $pos$ 已经出现了循环节，只需要在这中间看有没有停在 $0$ 的情况即可，$p$ 的范围是 $[1, 2n]$。 // Date: Wed Jan 3 23:12:16 2024 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; template <class T> using pq = priority_queue<T>; template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7; const ll INFL = 0x3f3f3f3f'3f3f3f3f; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nemp(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) #define NL cout << '\n'; template <class T> bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; } template <class T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; } template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } void __print(int x) { cerr << x; } void __print(long x) { cerr << x; } void __print(long long x) { cerr << x; } void __print(unsigned x) { cerr << x; } void __print(unsigned long x) { cerr << x; } void __print(unsigned long long x) { cerr << x; } void __print(float x) { cerr << x; } void __print(double x) { cerr << x; } void __print(long double x) { cerr << x; } void __print(char x) { cerr << '\'' << x << '\''; } void __print(const char *x) { cerr << '\"' << x << '\"'; } void __print(const string &x) { cerr << '\"' << x << '\"'; } void __print(bool x) { cerr << (x ? "true" : "false"); } template <typename T, typename V> void __print(const pair<T, V> &x) { cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}'; } template <typename T> void __print(const T &x) { int f = 0; cerr << '{'; for (auto &i : x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}"; } void _print() { cerr << "]\n"; } template <typename T, typename... V> void _print(T t, V... v) { __print(t); if (sizeof...(v)) cerr << ", "; _print(v...); } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #define dbg(x...) \ cerr << "\e[91m" << __func__ << ":" << __LINE__ << " [" << #x << "] = ["; \ _print(x); \ cerr << "\e[39m" << endl; #else #define debug1 #define debug2 #define debug3 #define dbg(x...) #endif void solve() { int n, x, p; cin >> n >> x >> p; int n1 = min(2 * n, p); For1(i, 1, n1) { if ((x + (i * ll(i + 1) / 2)) % n == 0) { cout << "YES\n"; return; } } cout << "NO\n"; } int main(void) { #ifdef _DEBUG freopen("1804c.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int T = 1; cin >> T; while (T--) { solve(); } return 0; }

Row Major tutorial 设连续的数字 $1, 2, \ldots, c$ 都能整除 $n$，并且 $c + 1 \nmid n$，那么位置为 $1, 2, \ldots, c, c+1$ 的字符都有可能相邻，并且它们任意两个位置都可能相邻。所以答案至少是 $c + 1$，取 $c+1$ 个不同的字母，然后不断重复这 $c + 1$ 个字符，可以发现相同字母的距离是 $k(c + 1)$，因为 $(c + 1) \nmid n$，所以 $k(c + 1) \nmid n$，因此相同字母不可能相邻，所以这个构造满足题意。 // Date: Tue Jan 2 22:51:33 2024 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; template <class T> using pq = priority_queue<T>; template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7; const ll INFL = 0x3f3f3f3f'3f3f3f3f; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nemp(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) #define NL cout << '\n'; template <class T> bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; } template <class T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; } template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } void __print(int x) { cerr << x; } void __print(long x) { cerr << x; } void __print(long long x) { cerr << x; } void __print(unsigned x) { cerr << x; } void __print(unsigned long x) { cerr << x; } void __print(unsigned long long x) { cerr << x; } void __print(float x) { cerr << x; } void __print(double x) { cerr << x; } void __print(long double x) { cerr << x; } void __print(char x) { cerr << '\'' << x << '\''; } void __print(const char *x) { cerr << '\"' << x << '\"'; } void __print(const string &x) { cerr << '\"' << x << '\"'; } void __print(bool x) { cerr << (x ? "true" : "false"); } template <typename T, typename V> void __print(const pair<T, V> &x) { cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}'; } template <typename T> void __print(const T &x) { int f = 0; cerr << '{'; for (auto &i : x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}"; } void _print() { cerr << "]\n"; } template <typename T, typename... V> void _print(T t, V... v) { __print(t); if (sizeof...(v)) cerr << ", "; _print(v...); } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #define dbg(x...) \ cerr << "\e[91m" << __func__ << ":" << __LINE__ << " [" << #x << "] = ["; \ _print(x); \ cerr << "\e[39m" << endl; #else #define debug1 #define debug2 #define debug3 #define dbg(x...) #endif void solve() { int n; cin >> n; string s; if (n == 1) { cout << "a\n"; } else if (n == 2) { cout << "ab\n"; } else { int c = 1; while (n % c == 0) c++; int cur = 0; For(i, 0, n) { s += ('a' + cur); cur = (cur + 1) % c; } cout << s << '\n'; } } int main(void) { #ifdef _DEBUG freopen("1844d.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int T = 1; cin >> T; while (T--) { solve(); } return 0; }

Mathematical Problem 用函数 $\_solve$ 把长度为 $1, 3, 5, 7$ 满足题目要求的数字打表打出来，能够发现频繁出现了 $1, 6, 9$ 和 $9, 6, 1$，只是中间添加了一些 $0$，原因是 $13^2$ 和 $31^2$ 计算的过程中没有进位，因此 $103^2$ 和 $301^2$ 在计算的过程中也没有进位。同时考虑到尾部添加偶数个 $0$ 也满足要求，再加上 $196$ 就可以凑够长度 $n$。 // Date: Sun Dec 31 11:33:41 2023 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; template <class T> using pq = priority_queue<T>; template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7; const ll INFL = 0x3f3f3f3f'3f3f3f3f; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nemp(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) #define NL cout << '\n'; template <class T> bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; } template <class T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; } template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } void __print(int x) { cerr << x; } void __print(long x) { cerr << x; } void __print(long long x) { cerr << x; } void __print(unsigned x) { cerr << x; } void __print(unsigned long x) { cerr << x; } void __print(unsigned long long x) { cerr << x; } void __print(float x) { cerr << x; } void __print(double x) { cerr << x; } void __print(long double x) { cerr << x; } void __print(char x) { cerr << '\'' << x << '\''; } void __print(const char *x) { cerr << '\"' << x << '\"'; } void __print(const string &x) { cerr << '\"' << x << '\"'; } void __print(bool x) { cerr << (x ? "true" : "false"); } template <typename T, typename V> void __print(const pair<T, V> &x) { cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}'; } template <typename T> void __print(const T &x) { int f = 0; cerr << '{'; for (auto &i : x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}"; } void _print() { cerr << "]\n"; } template <typename T, typename... V> void _print(T t, V... v) { __print(t); if (sizeof...(v)) cerr << ", "; _print(v...); } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #define dbg(x...) \ cerr << "\e[91m" << __func__ << ":" << __LINE__ << " [" << #x << "] = ["; \ _print(x); \ cerr << "\e[39m" << endl; #else #define debug1 #define debug2 #define debug3 #define dbg(x...) #endif bool check(ll x) { ll k = sqrtl(x); return k * k == x; } void solve() { int n; cin >> n; if (n == 1) { cout << "1\n"; return; } int k = (n - 3) / 2; For1(i, 0, k) { string y = string(i, '0'), y2 = string(n - 3 - i * 2, '0'); cout << "1" + y + "6" + y + "9" + y2 << '\n'; cout << "9" + y + "6" + y + "1" + y2 << '\n'; } cout << "196" + string(n - 3, '0') << '\n'; } void _solve() { int n; cin >> n; map<string, vector<ll>> m; For1(i, 1, 100000) { ll j = ll(i) * i; string js = to_string(j); sort(all(js)); m[js].pb(j); } for (auto &[s, vc] : m) { int len = SZ(s); if (SZ(vc) >= len && (len & 1) && (len == n)) { cout << "len = " << len << " : " << vc << '\n'; } } } int main(void) { #ifdef _DEBUG freopen("d.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int T = 1; cin >> T; while (T--) { solve(); } return 0; }

Training Before the Olympiad 发现只需要考虑数组中的奇数元素，每次 $B$ 操作最多会让结果减一，那么 $A$ 操作的目的是尽量减少奇数的个数，每次操作选择两个奇数操作就可以消除这两个奇数。$B$ 操作的时候尽量选择一个奇数和一个偶数，就可以尽量让结果变小。因此每一轮操作最多消除掉三个奇数，同时让结果比最优结果少 $1$，最终如果剩下 $1$ 个奇数，那么结果还要再减一。 // Date: Sun Dec 31 00:50:35 2023 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; template <class T> using pq = priority_queue<T>; template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7; const ll INFL = 0x3f3f3f3f'3f3f3f3f; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nemp(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) #define NL cout << '\n'; template <class T> bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; } template <class T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; } template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } void __print(int x) { cerr << x; } void __print(long x) { cerr << x; } void __print(long long x) { cerr << x; } void __print(unsigned x) { cerr << x; } void __print(unsigned long x) { cerr << x; } void __print(unsigned long long x) { cerr << x; } void __print(float x) { cerr << x; } void __print(double x) { cerr << x; } void __print(long double x) { cerr << x; } void __print(char x) { cerr << '\'' << x << '\''; } void __print(const char *x) { cerr << '\"' << x << '\"'; } void __print(const string &x) { cerr << '\"' << x << '\"'; } void __print(bool x) { cerr << (x ? "true" : "false"); } template <typename T, typename V> void __print(const pair<T, V> &x) { cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}'; } template <typename T> void __print(const T &x) { int f = 0; cerr << '{'; for (auto &i : x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}"; } void _print() { cerr << "]\n"; } template <typename T, typename... V> void _print(T t, V... v) { __print(t); if (sizeof...(v)) cerr << ", "; _print(v...); } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #define dbg(x...) \ cerr << "\e[91m" << __func__ << ":" << __LINE__ << " [" << #x << "] = ["; \ _print(x); \ cerr << "\e[39m" << endl; #else #define debug1 #define debug2 #define debug3 #define dbg(x...) #endif const int N = 100010; int a[N], n; void solve() { cin >> n; int odd = 0; ll sum{}; For1(i, 1, n) { cin >> a[i]; sum += a[i]; if (a[i] & 1) odd++; if (i == 1) { cout << a[i] << ' '; continue; } int mk = odd / 3, mo = odd % 3; ll res = sum - mk; if (mo == 1) res--; cout << res << ' '; } NL; } int main(void) { #ifdef _DEBUG freopen("c.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int T = 1; cin >> T; while (T--) { solve(); } return 0; }

Romantic Glasses 找到一个区间内的奇数下标元素的和与偶数下标元素的和相等，由于所有元素都为正数，可以把所有的奇数下标的值变成负数，这样如果某个区间的和是 0，就说明区间内奇数和偶数下标的和互为相反数并且绝对值相等。任意区间的和可以使用前缀和求。 // Date: Fri Dec 29 00:21:50 2023 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nonempty(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #else #define debug1 #define debug2 #define debug3 #endif const int N = 200010; int t, a[N], n; ll sum[N]; int main(void) { #ifdef _DEBUG freopen("e.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cin >> t; while (t--) { cin >> n; sum[0] = 0; set<ll> vis; bool flag = false; For1(i, 1, n) { cin >> a[i]; if (i & 1) a[i] = -a[i]; ll tmp = sum[i - 1] + a[i]; if (!flag && (has(vis, tmp) || tmp == 0)) { flag = true; } sum[i] = tmp; vis.insert(tmp); } cout << (flag ? "YES" : "NO") << '\n'; } return 0; }

Jellyfish and Green Apple tutorial 只需要考虑 $n < m$ 的情况，因为要求每个人得到的大小一样，所以每次切割都需要对现有的 $n$ 个全部切一次，也就是每次相当于 $n = 2n$，最终要满足 $m | (n \cdot 2^{k})$，要满足这一点，要求 $m$ 的除了 $2$ 之外的质因子，在 $n$ 中也要有，并且 $m$ 中除了 $2$ 的质因子的乘积能够整除 $n$ 中除了 $2$ 的质因子的乘积。这意味着 $\frac{m}{gcd(m, n)}$ 的质因数只有 $2$，也就是它是 $2$ 的幂次。可以利用这一点来判断是否有解。 接下来考虑最小的操作数。要满足 $m | (n \cdot 2^{k})$，而 $m \le 10^9 < 2^{30}$，所以最多操作 $30$ 次。为了让操作次数最小，每次 $n = 2n$ 之后，如果 $n > m$，那么 $n = n \mod m$，也就是每次都把能平均分的苹果都分下去，只留下不能平均分的，这样可以保证最终 $n = m$，同时每次操作 $n$ 的贡献都小于 $m$。 ...

Add, Divide and Floor tutorial 因为操作不会影响元素的顺序，所以可以把数组排序，经过一系列操作，使得最小值和最大值相等。设最小值和最大值分别是 $a, b$，考虑 $b - a$ 的值经过一次操作后的变化 $\lfloor\frac{b + x}{2}\rfloor - \lfloor \frac{a + x}{2} \rfloor$，枚举 $a, b$ 的奇偶性，观察发现 $b - a$ 最好情况下变成 $\frac{b - a}{2}$，当 $a$ 为奇数时，令 $x = 1$，否则令 $x = 0$。 // Date: Mon Dec 25 23:28:12 2023 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nonempty(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #else #define debug1 #define debug2 #define debug3 #endif const int N = 200010; int t, n, a[N]; int main(void) { #ifdef _DEBUG freopen("1901c.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cin >> t; while (t--) { cin >> n; int miv = INF, mav = -INF; For1(i, 1, n) { cin >> a[i]; miv = min(miv, a[i]); mav = max(mav, a[i]); } VI res; while (miv != mav) { if (miv & 1) { res.pb(1); miv = (miv + 1) / 2; mav = (mav + 1) / 2; } else { res.pb(0); miv /= 2; mav /= 2; } } int len = SZ(res); cout << len << '\n'; if (len <= n && len) { cout << res << '\n'; } } return 0; }

数学尖子生 先找到能够被 $1, 2, 3, \ldots, a - 1$ 整除的数字的个数，也就是求能够被 $lcm(1, 2, 3, \ldots, a - 1)$ 整除的数字的个数。同时要求不能被 $lcm(1, 2, 3, \ldots, a - 1, a)$ 整除，可以发现这两部分是子集的关系，直接相减即可。需要注意计算 $lcm$ 的时候，只需要计算 $lcm$ 在 $10^{16}$ 以内的，并且需要预处理。 // Date: Sun Dec 24 16:11:22 2023 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nonempty(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #else #define debug1 #define debug2 #define debug3 #endif const int N = 10000'10; ll b[N]; ll t, a, n; ll gcd(ll x, ll y) { return y == 0 ? x : gcd(y, x % y); } ll lcm(ll x, ll y) { return x / gcd(x, y) * y; } int main(void) { #ifdef _DEBUG freopen("e.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cin >> t; int n1 = 10000'10; b[1] = 1; For1(i, 2, n1) { if (b[i - 1] > 10000'0000'0000'0000) { break; } b[i] = lcm(b[i - 1], i); } while (t--) { cin >> a >> n; if (a == 1) { cout << "0\n"; continue; } ll k = b[a - 1], k1 = b[a], ans = 0; if (k && k1) ans = n / k - n / k1; cout << ans << '\n'; } return 0; }

取余 对于给定的 $b_i$，考虑 $a_j \mod b_i$ 的结果，其中 $a_j \in [1, a]$，可以发现当 $a_j < b_i$ 时，$a_j \mod b_i = a_j$；当 $a_j > b_i$ 时，存在循环节，可以求出循环节的个数和两端的剩余部分。对这三部分分别求出有多少个值在范围 $[S, T]$ 中，这是可以 $O(1)$ 求出的。 // Date: Sun Dec 24 15:30:01 2023 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nonempty(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #else #define debug1 #define debug2 #define debug3 #endif int a, b, s, t; ll cal(int l, int r) { if (l > r) return 0; int left = max(l, s), right = min(r, t); if (left > right) return 0; return right - left + 1; } ll get(int x) { ll res = 0; if (a < x) { return cal(1, a); } res += cal(1, x - 1); int m = (a - (x - 1)) / x; res += m * cal(0, x - 1); int k = (a - (x - 1)) % x; if (k) res += cal(0, k - 1); return res; } int main(void) { #ifdef _DEBUG freopen("d.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); while (cin >> a >> b >> s >> t) { ll res = 0; For1(i, 1, b) { res += get(i); } cout << res << '\n'; } return 0; }

Forever Winter tutorial 统计每个点的度数，难点在于边界条件。 degree count $x$ 1 1 $xy$ $y+1$ $x$ 边界条件是 $x = y + 1$ // Date: Fri Dec 22 12:15:15 2023 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nonempty(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #else #define debug1 #define debug2 #define debug3 #endif const int N = 210, M = 50010; int t, n, m, h[N], d[N]; int idx, e[M], ne[M]; void Init() { idx = 0; memset(d, 0, sizeof d); memset(h, -1, sizeof h); } void Add(int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; } int main(void) { #ifdef _DEBUG freopen("1829f.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cin >> t; while (t--) { cin >> n >> m; Init(); For(i, 0, m) { int x, y; cin >> x >> y; Add(x, y); d[x]++; d[y]++; } map<int, int> ma; For1(i, 1, n) { ma[d[i]]++; } int resx = -1, resy = -1; for (auto &[de, cnt] : ma) { if (cnt == 1) { resx = de; } } if (resx != -1) { for (auto &[de, cnt] : ma) { if (cnt == resx) { resy = de - 1; break; } } } else { for (auto &[de, cnt] : ma) { if (de != 1) { resy = de - 1; resx = cnt - 1; break; } } } cout << resx << ' ' << resy << '\n'; } return 0; }