A problem is easy

亦凉 2022-06-02 11:21 196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

[苗栋栋][Link 1]

思路：因为i+j+i\*j=n，所以变形为i+1+j+i\*j=n+1，所以为（i+1）\*（j+1）=（n+1）

代码：

#include<stdio.h>
    #include<math.h>
    #include<string.h>
    #include<stdlib.h>
    #include<algorithm>
    using namespace std;
    #define maxn 0x3f3f3f
    int main()
    {
        int test;
        scanf("%d",&test);
        while(test--)
        {
            int i,ans=0,n;//不知道为什么这里用long long就会超时
            scanf("%d",&n);
            for(i=1;(i+1)*(i+1)<=n+1;i++)
            {
                if((n+1)%(i+1)==0)
                    ans++;
            }
            printf("%d\n",ans);
        }
        return 0;
    }

[C154-1002-1.jpg]: /images/20220602/88eb8da2f5cb40a08f5c7ad8bdf5952e.png
[Link 1]: http://acm.nyist.edu.cn/JudgeOnline/profile.php?userid=%E8%8B%97%E6%A0%8B%E6%A0%8B