对于每一行,从大到小放是收益最大的。由于每个矩形的长度都是 2 的幂,所以从大到小循环 2 的幂即可,直到把所有的矩形都放完。
// Date: Sun Dec 10 12:25:53 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 = 100010;
int t, n, w, a[N];
int main(void) {
#ifdef _DEBUG
freopen("1498b.in", "r", stdin);
#endif
std::ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> t;
while (t--) {
cin >> n >> w;
map<int, int> m;
For1(i, 1, n) {
cin >> a[i];
m[a[i]]++;
}
int res = 0;
while (nonempty(m)) {
res++;
int rem = w;
int len = SZ(m);
Rof1(i, 0, 31) {
int x = (1 << i);
if (!rem)
break;
if (!has(m, x) || rem < x)
continue;
int cnt = m[x];
int hold = rem / x;
if (hold >= cnt) {
m.erase(x);
rem -= cnt * x;
} else {
m[x] -= hold;
if (!m[x])
m.erase(x);
rem -= hold * x;
}
}
}
cout << res << '\n';
}
return 0;
}