B - An easy problem

本文主要是介绍B - An easy problem，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

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

输入n，计算有多少种i*j+i+j==n

#include <iostream>
#include <cstring>
#include <cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
int main()
{int t;scanf("%d",&t);long long n;for(int k=0;k<t;k++){cin>>n;n++;int m=(int)sqrt(n*1.0);int count=0;for(int i=2;i<=m;i++){if(n%i==0)count++;}cout<<count<<endl;}return 0;
}

 

这篇关于B - An easy problem的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---