Modular Inverse ZOJ

题目传送门

题意：ax≡1 (mod m) 求x

思路：就是一个裸的扩展欧几里得，但是要注意一下m为1的情况

#include <algorithm>#include <cmath>#include <cstdio>#include <cstring>#include <fstream>#include <iostream>#include <list>#include <map>#include <queue>#include <set>#include <sstream>#include <stack>#include <string>#include <vector>#define MAXN 20100#define MAXE 5#define INF 1000000000#define MOD 9901#define LL long long#define pi acos(-1.0)using namespace std;void exgcd(LL a, LL b, LL &d, LL &x, LL &y) {    if (!b) {        d = a;        x = 1;        y = 0;    } else {        exgcd(b, a % b, d, y, x);        y -= x * (a / b);    }}int main() {    std::ios::sync_with_stdio(false);    int T;    cin >> T;    LL a, m;    for (int kase = 1; kase <= T; ++kase) {        cin >> a >> m;        if (m == 1) {            cout << 1 << endl;        } else {            LL d, x, y;            exgcd(a, m, d, x, y);            if (d != 1) {                cout << "Not Exist\n";            } else {                cout << (x + m) % m << endl;            }        }    }    return 0;}