A problem is easy
時間限制：1000 ms | 內存限制：65535 KB
難度：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 ?
輸入
The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10^11).
輸出
For each case, output the number of ways in one line
樣例輸入
2
1
3
樣例輸出
0
1
import java.io.BufferedInputStream;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner cin = new Scanner(new BufferedInputStream(System.in));
int T = cin.nextInt();
while(T-->0){
int num = cin.nextInt();
int N = 0;
for(int i=2;i<=Math.sqrt(num+1);i++){
if((num+1)%i==0)
N++;
}
System.out.println(N);
}
}
}