ZOJ 3609 Modular Inverse【扩展欧几里得】

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题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3609

思路：

(ax-1)%m == 0 -> ax-mk = 1; 求 x

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<queue>
#include<map>
#include<stack>
#include<sstream>
#include<vector>
#include<string>
#include<set>

using namespace std;

#define IOS ios::sync_with_stdio(false); cin.tie(0);

int read(){
    int r=0,f=1;char p=getchar();
    while(p>'9'||p<'0'){if(p=='-')f=-1;p=getchar();}
    while(p>='0'&&p<='9'){r=r*10+p-48;p=getchar();}return r*f;
}

typedef long long ll;
const int Maxn = 30010;
const long long LINF = 1e18;
const int INF = 0x3f3f3f3f;

void exgcd (int a,int b,int &d,int &x, int &y) {
    if(!b) { d = a; x = 1; y = 0; }
    else {
        exgcd (b,a%b,d,y,x); y-=(a/b)*x;
    }
}

int main (void)
{
    int t,a,m,ans;
    cin >> t;
    while (t--) {
        cin >> a >> m;
        int xx,yy,g;
        exgcd(a,m,g,xx,yy);  // xx为a的系数，yy为m的系数，g是gcd（a，m）
        if(g == 1) {
            while (xx <= 0) xx+=m;  // y的通解 y0-k*a;  x的通解  x0+k*b; k为任意整数，y和x统一一个k
            cout << xx << endl;
        }
        else cout << "Not Exist" << endl;
    }
    return 0;
}