南阳题目216-A problem is easy

﹏ヽ暗。殇╰゛Y 2022-07-28 05:10 159阅读 0赞

## 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..  
![C154-1002-1.jpg][]  
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

这是一个数学题目，找出规律就可以了

我们看题目要求，N= i \* j + i + j (0 < i <= j)

我们让等式两边同时加以1 i\*(j+1)+j+1=N+1

所以 (i+1)\*(j+1)=N+1

所以只要是N+1可以整除j即可

j的取值范围就是1-sqrt（N）-1

所以计算个数就行了

#include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    int main()
    {
    	int a,b,c,d,i,j,m,n,sum;
    	scanf("%d",&m);
    	while(m--)
    	{
    		sum=0;
    		scanf("%d",&n);
    		for(i=1;i<=sqrt(n)-1;i++)
    		{
    			if((n+1)%(i+1)==0)
    			sum++;
    		}
    		printf("%d\n",sum);
    	}
    	return 0;
    }

[C154-1002-1.jpg]: /images/20220727/d48e9fe5541a42a4aa4b70a480982631.png