HDU - 2601 An easy problem

最新推荐文章于 2019-08-07 19:04:40 发布

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博客描述了一个有趣的数学问题，即给定一个数N，求解有多少种方式可以表示为i*j+i+j的形式（0<i<=j）。通过巧妙的数学转换，问题转化为求解N+1的因子对数。代码实现使用C++，通过遍历至平方根的方法高效求解。

An easy problem

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10698 Accepted Submission(s): 2728

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 <= 10^10).

Output

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

Sample Input

2 1 3

Sample Output

0 1

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

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题意：给n 求满足i*j+i+j=n(0<i<=j)方案数

思路：xy+x+y=n
(x+1)(y+1)=x*y+x+y+1=n+1;

即求n+1的因子对数

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
int main()
{
	int t;scanf("%d",&t);
	while(t--)
	{
		LL n;
		scanf("%lld",&n);
		n++;
		int m=sqrt(n);
		int ans=0;
		for(int i=2;i<=m;i++)
		{
			if(n%i==0)ans++;
		}
		printf("%d\n",ans);
	}
}