【ZOJ 3609】Modular Inverse 最小乘法逆元

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

题解：
最小乘法逆元：由ax≡1 (mod m)得：转化为解线性方程ax+by=1
需要注意的地方：最小解取模时不能写成（x%t+t）%t 因为此题要的是正数解 这样写有时会输出0

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 typedef long long ll;
 ll exgcd(ll a,ll b,ll &x,ll &y)
 {
     if(!b)
     {
         x=;y=;
         return a;
     }
     ll r=exgcd(b,a%b,x,y);
     ll t=x;
     x=y;
     y=t-a/b*y;
     return r;
 }
 void work(ll a,ll b,ll c)
 {
     ll x,y;
     ll r=exgcd(a,b,x,y);
     if(c%r!=){
         printf("Not Exist\n");
         return ;
     }
     x*=c/r;
     ll t=b/r;
     if(t<)t=-t;
     x%=t;
     if(x<=)x+=t;
     printf("%lld\n",x);
 }
 int main()
 {
     int T;
     ll a,b;
     scanf("%d",&T);
     while(T--){
         scanf("%lld%lld",&a,&b);
         work(a,b,);
     }
     return ;
 }