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 + );

         long long cnt = ;

         for (int i = ; i <= b; ++ i)

         {

             if ((a + ) % i == )

                 ++ cnt;

         }

         printf("%lld\n", cnt);

     }

 }