Modular Inverse(模逆元，扩展欧几里德)

Modular Inverse

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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：求最小正整数解，其实吧，x的通解是x0+b/gcd*t，由于t是整数，那么ans=x0+b/gcd*t=x0 mod b=x0%b；因为ans要是正整数的，
所以当b/gcd是负的时候，就等于绝对值就好了，因为还有t啊，当x0%b负时,加上一个b；就妥了；因为ans=(x0+b)%b;
代码：

 #include<iostream>
 #include<algorithm>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 using namespace std;
 const int INF=0x3f3f3f3f;
 typedef long long LL;
 void e_gcd(LL a,LL b,LL &d,LL &x,LL &y){
     if(!b){
         d=a;
         x=;
         y=;
     }
     else{
         e_gcd(b,a%b,d,x,y);
         LL temp=x;
         x=y;
         y=temp-a/b*y;
     }
 }
 LL cal(int a,int b,int c){
     LL x,y,d;
     e_gcd(a,b,d,x,y);
     if(c%d!=)return -;//ax+by=c/(c/gcd);
     x*=c/d;
      b/=d;//因为x的通解是x0+(b/gcd)t;
      if(b<)b=-b;
      LL ans=x%b;
      if(ans<=)ans+=b;
      return ans;
 }
 int main(){
     LL T,a,b,d,x,y;
     scanf("%d",&T);
     while(T--){
         scanf("%lld%lld",&a,&b);
         LL ans=cal(a,b,);
         if(ans==-)puts("Not Exist");
         else printf("%lld\n",ans);
     }
     return ;
 }

题上数据比较水，数据范围1000，暴力一下就可以了：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int INF=0x3f3f3f3f;
typedef long long LL;
int main(){
    int T,a,m;
    scanf("%d",&T);
    while(T--){//(1-ax)%m;
        scanf("%d%d",&a,&m);
        int flot=;
        for(int x=;x<=;x++){
            if((-a*x)%m==){
                flot=;
                printf("%d\n",x);
                break;
            }
        }
        if(flot)continue;
            puts("Not Exist");
    }
    return ;
}