bzoj 1477 扩展欧几里德

思路：很裸的求相遇问题。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N = 2e6 + ;
const int M = 1e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +;

LL a, b, c, m, n, L, x, y;

LL exgcd(LL a, LL b, LL &x, LL &y) {
    if(!b) {
        x = ; y = ;
        return a;
    } else {
        LL gcd, t; gcd = exgcd(b, a % b, x, y);
        t = x; x = y; y = t - (a / b) * y;
        return gcd;
    }
}

int main() {
    scanf("%lld%lld%lld%lld%lld", &a, &b, &m, &n, &L);
    LL gcd = exgcd(n - m, L, x, y);
    if((a - b) % gcd != ) {
        puts("Impossible");
    } else {
        x *= ((a - b) / gcd);
        x %= L / gcd;
        if(x < ) x += abs(L / gcd);
        printf("%lld\n", x);
    }
    return ;
}

/*
*/