An easy problem 2601 （数学题+公式变形）

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没有能与不能只有想与不想

最新推荐文章于 2025-01-21 23:56:51 发布

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分类专栏： hdoj 数学----------------------------

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本文探讨了Teddy遇到的一个复杂数学问题——如何计算N个元素的特定组合方式。通过鲁莱先生的指引，Teddy发现了解决方法，并在代码实现中展示了求解过程。文章提供了一个输入输出示例，帮助读者理解问题和解决方案。

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An easy problem

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8060 Accepted Submission(s): 1959

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
 
//题中的公式可以变形为 n=(i+1)(j+1)-1;==>>(n+1)=(i+1)(j+1);
//这样就可以了
 
#include<stdio.h>
#include<string.h>
#include<math.h>
int main()
{
	int t,m;
	long long n;
	int i,j;
	scanf("%d",&t);
	while(t--)
	{
		m=0;
		scanf("%lld",&n);
		for(i=2;i<=sqrt(n+1);i++)
		{
			if((n+1)%i==0)
				m++;
		}
		printf("%d\n",m);
	}
	return 0;
}