NYOJ 216 A problem is easy

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

双重循环肯定超时的 直接用一个判断条件能减少时间

i*j+i+j =N 经过观察，可以变形为i*j+i+j+1=N+1,也就是说，可以进一步变形为(i+1)*(j+1)=N+1

所以i从2判断到sqrt(n+1)即可

#include<stdio.h>
main(){
int n,i;
scanf("%d",&n);
while(n--){
int num,sum=0;
scanf("%d",&num);
for(i=2;i*i<=num+1;i++){
if((num+1)%i==0)
sum++;
}
printf("%d\n",sum);
}

}