An easy problem+数学推理

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本文介绍了一个有趣的数学问题：对于给定的N，如何计算可以用i*j+i+j形式表示的方法数。通过数学推理的方式解决了这个问题，并给出了C语言实现的代码示例。

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An easy problem

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4575 Accepted Submission(s): 1117

Problem Description

When Teddy was a child , he was always thinking about some simple math problems ,such as “What it’s 1 cup of water plus 1 pile of dough ..” , “100 yuan buy 100 pig” .etc..

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 <= 10¹⁰).

Output

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

Sample Input

2
1
3

Sample Output

0
1
思路：数学推理
 
#include<stdio.h>
#include<math.h>
int main()
{
	int n,k;
	__int64 i,j,m;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%I64d",&m);
		m=m+1;
		j=sqrt(m);
		k=0;
			for(i=2;i<=j;i++)
			{
				if(m%i==0)
				k++;
			}
		printf("%d\n",k);
	}
	return 0;
}