思路：二分K，查看当前K能否满足总时间不超过最大时间

ACcode:

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e4+10;
int n,tmax,a[N];
bool check(int x) {
	priority_queue < int, vector < int >, greater < int > > q;
	int y=0,ans=0;
	for(int i=1; i<=x; i++) q.push(a[i]);
	for(int i=x+1; i<=n; i++) {
		ans+=q.top()-y;
		if(ans>tmax)return false;//小剪枝
		y=q.top();
		q.pop();
		q.push(a[i]+y);
	}
	for(int i=1; i<=x; i++) {
		ans+=q.top()-y;
		if(ans>tmax)return false;//小剪枝
		y=q.top();
		q.pop();
	}
	return ans<=tmax;
}
void solve() {
	cin>>n>>tmax;
	for(int i=1; i<=n; i++) cin>>a[i];
	int l=0,r=n+1;
	while(l+1<r) {
		int mid=l+r>>1;
		if(check(mid))r=mid;
		else l=mid;
	}
	cout<<r<<"\n";
}
signed main() {
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	solve();
	return 0;
}

注释：

over~