POJ 2853

#include<iostream>
#include<stdio.h>
#include<vector>
#include<math.h>
#include<algorithm>
#define FOR for
#define M 10009
using namespace std;
int main()
{
	//freopen("acm.acm","r",stdin);
	unsigned sum;
	unsigned  num;
	unsigned  ans;
	unsigned pos;
	unsigned  pos1;
	int time;
	int tem;
	int i;
	cin>>time;
	while(time --)
	{
		ans = 0;
		cin>>tem>>num;
		for(i = 1; i < sqrt(long double(num*2)); ++ i)
		{
			if((2 * num % i == 0) && (i + 2 * num / i) % 2 != 0)
			{
				++ ans;
			}
		}
		cout<<tem<<" "<<--ans<<endl;
	}
}

/*
需求出这个不定方程的解数即可。显然2 * i + j - 1 > j,于是j < sqrt(2 * n),
方程有解的条件是(2 * n % j == 0) && (j + 2 * n / j) % 2 != 0 有多少个ｊ满足条件就有多少个解，编程实现就很简单了
*/