南阳oj入门题-a problem is easy

最新推荐文章于 2022-11-17 19:36:24 发布

原创最新推荐文章于 2022-11-17 19:36:24 发布 · 500 阅读

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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，求有多少种方式可以将N表示为i*j+i+j的形式，其中0<i<=j。通过分析，我们提供了一种有效的算法来解决这个问题，特别适用于当N较大时的场景。

/**
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<iostream>
#include<math.h>
using namespace std;
int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		int n,sum=0;
		cin>>n;
		for(int i=2;i<=sqrt(n+1);i++)
			if((n+1)%i==0)
					sum++;
		cout<<sum<<endl;
	}
	return 0;
 }