NYOJ - A problem is easy

最新推荐文章于 2018-09-08 23:34:00 发布

原创最新推荐文章于 2018-09-08 23:34:00 发布 · 639 阅读

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本文介绍了一个有趣的数学问题，即给定一个数N，求出有多少种方式可以将N表示为i*j+i+j的形式（0<i≤j）。文章提供了一段C语言实现的代码，通过该算法能够有效地解决这个问题。

A problem is easy

时间限制：1000 ms | 内存限制：65535 KB

难度：3

描述

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 ?

输入

The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10^11).

输出

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

样例输入

2
1
3

样例输出

0
1

#include <stdio.h>
#include <math.h>
int main()
{
	int T;
	int  i,j,n;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		int s = 0;
		int k = (int)sqrt(n+1);
		for(i = 2; i <= k; ++i)
		{
			if((n+1) % i == 0)
			s++;
		}
		printf("%d\n",s);
	}
}