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An easy problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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Total Submission(s): 9321    Accepted Submission(s): 2301

Problem 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 <= 10 ¹⁰).

Output

For each case, output the number of ways in one line.

Sample Input

2 1 3

Sample Output

0 1

思路：注意到 ( i +1 ) * ( j + 1 ) = i * j + i + j + 1= n + 1；而且这里的 n 比较大，只能跑一层循环（杭电上能过的题，去 nyoj 上就 TEL，也是服了）

#include<cstdio>
#include<cstring>
#include<cmath> 
#define LL long long
using namespace std;
LL n;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld",&n);
        int ans=0;
        for(LL i=2;i*i<=n+1;i++)
        {
            if((n+1)%i==0)
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}