Codeforces 495 B. Modular Equations && Codeforces Round #282 (Div. 2)

题目链接：

http://codeforces.com/problemset/problem/495/B

解题思路：

Codeforces官方题解：

• If a < b then there is no answer since .
• If a = b then x can be any integer larger than a. so there are infinite number of answers to the equation.
• The only remaining case is when a > b. Suppose x is an answer to our equation. Then x|a - b. Also since  then b < x. These conditions are necessary and sufficient as well. So the answer is number of divisors of a - b which are strictly greater than bwhich can be solved in .

 当a==b时，直接输出“infinity”。不能直接遍历（O（a-b）），会超时，需要开放求解。

AC代码：

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    int a,b;
    while(scanf("%d%d",&a,&b)!=EOF)
    {
        int i,sum=0;
        if(a==b)
        {
            printf("infinity\n");
            continue;
        }
        int t=a-b;
        int j=sqrt(t*1.0);
        for(i=1;i<=j;i++)
        {
            if(t%i==0)
            {
                int tt=t/i;
                if(i>b)
                    sum++;
                if(tt!=i&&tt>b)
                    sum++;
            }
        }
        printf("%d\n",sum);

    }
    return 0;
}