NYOJ-216-A problem is easy

迷南。 2022-05-30 08:55 249阅读 0赞

A problem is easy

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..
C154-1002-1.jpg
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[sup]10[/sup]).

Output

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

Sample Input

2 1 3

Sample Output

0 1

  1. #include<stdio.h>
  2. int main()
  3. {
  4.     int i,j,s,m,sum;
  5.     scanf("%d",&s);
  6.     while(s--)
  7.     {
  8.         sum=0;
  9.         scanf("%d",&m);
  10.         for(i=1;(i+1)*(i+1)-1<=m;i++)
  11.         {
  12.             if((m+1)%(i+1)==0)
  13.                 sum++;
  14.         }
  15.         printf("%d\n",sum);
  16.     }
  17.     return 0;
  18. }

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