NYOJ - A problem is easy

最新推荐文章于 2022-10-17 01:11:50 发布

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本文探讨了Teddy在梦境中遇到老者鲁莱提出的一个数学问题，即如何计算给定数量N时，表达式asi+j(i<=j)的可能组合数。通过实例输入和输出，展示了解决此类问题的方法，旨在帮助读者理解并掌握复杂数学问题的求解策略。

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 n,m;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%d",&m);
		m++;
		int count = 0;
		for(int i = 1; i <= sqrt(m); i++)    // (i+1)*(j+1) = m+1
		{
			if(m % (i+1) == 0&&(m/(i+1)-1>0))
			count++;
		}
		printf("%d\n",count);
	}
}