HDOJ 题目2601 An easy problem（数学水题）

最新推荐文章于 2018-04-10 12:54:00 发布

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Teddy遇到了一个数学问题：对于给定的N，有多少种方式可以表示为N=i*j+i+j (0<i≤j)。通过将问题转化为求解(n+1)=(i+1)*(j+1)的形式，本文提供了一种有效的解决方法。

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An easy problem

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6196 Accepted Submission(s): 1490

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 <= 10¹⁰).

Output

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

Sample Input

2
1
3

Sample Output

0
1

Author

Teddy

思路：因式分解n=i*j+i+j,即（n+1）=(i+1)(j+1);

ac代码

#include<stdio.h>
#include<math.h>
int main()
{
	int i,j,n,t;
	scanf("%d",&t);
	while(t--)
	{
		__int64 n,nc;
		int c=0,i;
		scanf("%I64d",&n);
		n++;
		nc=sqrt(n);
		for(i=2;i<=nc;i++)
		{
			if(n%i==0)
				c++;
		}
		printf("%d\n",c);
	}
}