感觉思路没错啊

在最大和最小之间不就输出Yes 否则No嘛

贴一下只过Subtask#4的垃圾代码：

#include <bits/stdc++.h>
using namespace std;
int main()
{
	int T;
	cin >> T;
	while (T--)
	{
		int n, k, s, maxx = 0, minn = 0;
		cin >> n >> k >> s;
		maxx = (n - k + 1 + n) * k / 2;
		minn = (1 + 1 + k - 1) * k / 2;
		if (s <= maxx && s >= minn)
		{
			cout << "Yes" << endl;
		}
		else
		{
			cout << "No" << endl;
		}
	}
	return 0;
}