A problem is easy

A problem is easy

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 <cmath>
using namespace std;
int main()
{
    int n;
    int a,b;
    cin>>n;
    while(n--)
    {
        cin>>a;
        b=0;
        for(int i=2;i*i<=(a+1);i++)
        {
            int j = (a+1)/i;
            if(i*j==a+1 && i<=j)
        
                b++;
        }
        cout<<b<<endl;
    }
return 0;
}        

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