NYOJ-216-A problem is easy-2013年10月17日13:57:33

A problem is easy

1000 ms  |  内存限制： 65535

3

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 ?

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<stdio.h>

int main()
{
  int i,j,N,T,count;
  scanf("%d",&T);
  while(T--)
  {
    count = 0;
    scanf("%d",&N);
    for(j=2;j*j<=N+1;j++)
      if((N+1)%j==0)
        count++;
    printf("%d\n",count);

  }

  return 0;
}