nyoj 216-A problem is easy ((i + 1) * (j + 1) = N + 1)

216-A problem is easy

内存限制:64MB 时间限制:1000ms 特判: No
通过数:13 提交数:60 难度:3

题目描述:

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 ?

输入描述:

The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10^11).

输出描述:

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

样例输入:

复制

2
1
3

样例输出:

0
1

C/C++  AC:

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <stack>
#include <set>
#include <map>
#include <queue>
#include <climits>

using namespace std;

int main()
{
    long long T, a, b;
    cin >>T;
    while (T --)
    {
        scanf("%lld", &a);
        b = sqrt(a + 1);
        long long cnt = 0;
        for (int i = 2; i <= b; ++ i)
        {
            if ((a + 1) % i == 0)
                ++ cnt;
        }
        printf("%lld
", cnt);
    }
}