NYOJ 216 A problem is easy

最新推荐文章于 2019-02-28 19:03:35 发布

原创最新推荐文章于 2019-02-28 19:03:35 发布 · 823 阅读

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本文介绍了一道Teddy遇到的数学难题，并提供了一个简洁高效的算法解决方案。通过将原始问题转化为寻找特定因子的问题，该算法能在给定的时间复杂度内找到所有符合条件的组合。

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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

刚开始用了两层for循环，不管怎么改都超时了，看到讨论群，才解出来。

解题思路：i*j+i+j=n可以转化为i*j+i+j+1=n+1,即可以写成(i+1)*(j+1)=n+1; 便可以把i+1看成一个大于1的数，找出n+1的另一个因子就可以了

代码如下：

#include<stdio.h>
#include<stdlib.h>
int main()
{
	int t,n,i,count;
	scanf("%d",&t);
	while(t--)
	{
		count=0;
		scanf("%d",&n);
		for(i=2;i*i<=(n+1);i++)
		{
			if((n+1)%i==0)
			   count++;
		}
		printf("%d\n",count);
	}
	return 0;
}