Winter D3 D Modular Inverse

Modular Inverse Time limit: 2 Seconds Memory Limit: 65536 KB

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

Input

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

Each test case contains 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

33 114 125 13

Sample Output

4Not Exist8

#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;}      In qualifying, it's been here. WA dropped, remember:      //a three B (mod m) is the same as  : a mod m = b mod m      //Can be expressed as A=b+m*k  where K is starting from 0 for      (K=0;K&L t;=1000;k++)      {          if ((m*k+1)%a==0) {printf ("%d\n", (m*k+1)/a);   return 0;}

Winter D3 D Modular Inverse