zoj3609 Modular Inverse（扩展欧几里德算法求逆元）模板

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

题意：已知a，m求最小的x满足ax≡1（mod m）

代码：扩展欧几里得求逆元代码

#include <cstdio>

#include <iostream>

using namespace std;

int ext_gcd(int a,int b,int &x,int &y){//扩展欧几里得求,一组解x,y

    if(!b) {x=1,y=0;return a;}

    int d=ext_gcd(b,a%b,y,x);

    y-=a/b*x;

    return d;

}

int Inv(int a,int m){

    int d,x,y;

    d=ext_gcd(a,m,x,y);

    if(1!=d) return -1;//gcd(a,b)!=1此时无解，证明a,m不互质，

    else return (x%m+m)%m;//x可能为负，相当于取绝对值

}

int main()

{

    int T,a,m;

    cin>>T;

    while(T--)

    {

        cin>>a>>m;

        if(m==1){

            puts("1");

            continue;

        }

        int d=Inv(a,m);

        if(-1==d) puts("Not Exist");

        else cout<<d<<endl;

    }

    return 0;

}