A. Deadline （不等式、分块整除）

题目

分块整除⌈x+d/（x+1）​⌉=⌊x+(d+x)/(x+1)​⌋=1+⌊x+(d−1)/(x+1)​⌋

Code:

#include<iostream>
#include<cmath>
using namespace std;
typedef long long ll;
const int Max = 1e6 + 5;	
const int Mod = 1e9 + 7;

int main()
{
    
    
	int t;cin >> t;
	while (t--)
	{
    
    
		int n, d;cin >> n >> d;
		int f = 0;
		for (int l = 1, r;l <= d;l=r+1)
		{
    
    
			r = d /(d/l);
			if (l + (d-1)/ l <= n)
			{
    
    
				f = 1;break;
			}
		}
		if (f)cout << "YES" << endl;
		else cout << "NO" << endl;
	}

}

不等式

参考博客https://www.cnblogs.com/pixel-Teee/p/12199936.html

只需判断2*sqrt(d)-1是否<=n即可