A problem is easy-蒲公英云

A problem is easy

亦凉 2022-06-02 11:21 283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..

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

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;
}