Hdoj 2058

原题链接

描述

Given a sequence 1,2,3,......N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M.

输入

Input contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000000).input ends with N = M = 0.

输出

For each test case, print all the possible sub-sequence that its sum is M.The format is show in the sample below.print a blank line after each test case.

样例输入

20 10
50 30
0 0

样例输出

[1,4]
[10,10]

[4,8]
[6,9]
[9,11]
[30,30]

思路

等差数列求和的变形。
\(S=\frac{(k+k+n-1)*n}{2}\)
\(2k=\frac{2s}{n}-n+1, n≤\sqrt{S}\)
枚举n然后算出k就好

代码

#include <bits/stdc++.h>
#define ll long long
using namespace std;

int main()
{
	ll n, m;
	while(~scanf("%lld %lld", &n, &m))
	{
		if(n + m == 0) break;
		ll k, t;
		for(t = sqrt(2 * m); t > 0; t--)
		{
			if(2 * m % t) continue;
			ll r = 2 * m / t - t + 1;
			if(r % 2) continue;
			k = r >> 1;
			printf("[%lld,%lld]\n", k, k + t - 1);
		}
		printf("\n");
	}
	return 0;
}