An easy problem

最新推荐文章于 2020-03-09 17:27:11 发布

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最新推荐文章于 2020-03-09 17:27:11 发布

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原文链接：http://www.cnblogs.com/-xss/p/5943987.html

本文介绍了一个有趣的数学问题：如何计算对于给定的N值，有多少种方式可以表示为i*j+i+j的形式（0<i≤j）。通过分析并提供了一段C++代码实现，展示了如何有效地解决这个问题。

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One day Teddy met a old man in his dream , in that dream the man whose name was“RuLai” gave Teddy a problem :

Given an N , can you calculate how many ways to write N as i * j + i + j (0 < i <= j) ?

Teddy found the answer when N was less than 10…but if N get bigger , he found it was too difficult for him to solve.
Well , you clever ACMers ,could you help little Teddy to solve this problem and let him have a good dream ?

Input

The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10 ¹⁰).

Output

For each case, output the number of ways in one line.

Sample Input

2
1
3

Sample Output

0
1

#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		int ans=0;
		long long N;
		scanf("%lld",&N);
		/*for(int i=1;(i+1)*(i+1)<=(N+1);i++)
		{
			if(((N-i)%(i+1)==0)&&((N-i)/(i+1)>=i))
				ans++;
		}*/
		for(long long i=1;(i+1)*(i+1)<=(N+1);i++)
		{
			if((N+1)%(i+1)==0&&((N-i)/(i+1)>=i))
				ans++;
		}
		printf("%d\n",ans);
	}
	return 0;
 }