寒假 D3 D Modular Inverse

The modular modular multiplicative inverse of an integer a modulo m is an integer x such that a-1≡x (mod m). This is equivalent to ax≡1 (mod m).

Input

There are multiple test cases. The first line of input is an integer T ≈ 2000 indicating the number of test cases.

Each test case contains two integers 0 < a ≤ 1000 and 0 < m ≤ 1000.

Output

For each test case, output the smallest positive x. If such x doesn‘t exist, output "Not Exist".

Sample Input

3
3 11
4 12
5 13

Sample Output

4
Not Exist
8

#include<cstdio>
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
int main()
{
    int i,j,k;
    int a,m;
   int T;
   scanf("%d",&T);
   while(T--)
   {
       scanf("%d%d",&a,&m);
      if(gcd(a,m)!=1) {printf("Not Exist\n");continue;}
      //排位赛的时候一直在这里wa掉，谨记：
      //a三b（mod m) 是同余  即：a mod m == b mod m
      //可表示成 a=b+m*k  其中k是从 0 开始
      for(k=0;k<=1000;k++)
      {
          if((m*k+1)%a==0) {printf("%d\n",(m*k+1)/a);break;}
      }
   }
   return 0;
}

寒假 D3 D Modular Inverse

原文：http://www.cnblogs.com/orchidzjl/p/4307530.html