先找到所有 $a_i < i$ 的元素放到数组 $b$,数组根据 $a_i$ 排序,再对每个 $b_i$ 找到第一个大于 $i$ 的元素,可以使用 upper_bound
。另外, lower_bound
可以找到第一个 $\ge$ 给定元素的位置。
// Date: Sat Dec 16 09:52:26 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())
#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];
vector<PII> b;
int main(void) {
#ifdef _DEBUG
freopen("1703f.in", "r", stdin);
#endif
std::ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> t;
while (t--) {
cin >> n;
b.clear();
For(i, 0, n) {
cin >> a[i];
if (a[i] < i + 1) {
b.pb({a[i], i + 1});
}
}
ll res = 0;
if (b.empty()) {
cout << "0\n";
continue;
}
sort(all(b));
for (auto &[x, y] : b) {
auto it = upper_bound(b.begin(), b.end(), PII{y, INF});
if (it != b.end()) {
res += b.end() - it;
}
}
cout << res << '\n';
}
return 0;
}