nyoj.216 A problem is easy【水题】 2015/03/15

A problem is easy

时间限制：1000 ms | 内存限制：65535 KB
难度：3

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>
#include<stdlib.h>
// i*j+i+j=n  =>  (i+1)*(j+1)=n+1
int main(){
int t,n,i,num;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
num = 0 ;
for( i = 1 ; (i+1)*(i+1) <= (n+1) ; ++i )
if( (n+1) % (i+1) == 0 )
num++;
printf("%d\n",num);
}
return 0;
}

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