【HDU 2601 】

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 <= 1010).

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

Sample Input
2
1
3

Sample Output
0
1

不难想到 i * j + i +j = (i + 1) * (j + 1) - 1 = N,既N + 1 = （i + 1） * (j + 1),只用判断 N + 1 是否能整除 i + 1 就好因为 j >= i,所以只用判断到 根号 N +1就好

AC代码 ：

#include<cstdio>#include<cmath>using namespace std;int main(){    int T,i,ans;    long long sum,N;    scanf("%d",&T);    while(T--)    {        scanf("%lld",&N);        sum = N + 1;        ans = 0;        N = sqrt(sum * 1.0);        for(i = 2 ; i <= N; i++)            if(sum % i == 0)             ans++;        printf("%d\n",ans);    }    return 0;}