A problem is easy http://acm.nyist.net/JudgeOnline/problem.php?pid=216

 

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

上传者

苗栋栋

#include<stdio.h>
int main()
{
	int n;
	scanf("%d",&n);
	while(n--)
	{
		long int i,j,m,count=0;
		scanf("%d",&m);
		if(m==1||m==0)
			printf("%d
",0);
		else
		{
			for(i=1;i*i<=m;i++)
			{
				if((m+1)%(i+1)==0)
				{
					j=(m+1)/(i+1)-1;
					if(i<=j)
						count++;
				}
			}
			printf("%d
",count);
		}
	}
	return 0;
}

这一题不能用两个for循环遍历求解，那样会超时，我们来看一下，N=i*j+i+j,两边同时加1，N+1=i*(j+1)+j+1=(i+1)*(j+1);所以我们只要有（N+1）%(i+1)==0;就可以确定一个j，要注意的是i小于j.