HDU 5114 扩展欧几里得

题目大意：给你两个球的坐标 他们都往(1, 1)这个方向以相同的速度走，问你他们在哪个位置碰撞。

思路：这种题目需要把x方向和y方向分开来算周期，两个不同周期需要用扩展欧几里得来求第一次相遇。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg

using namespace std;

const int N = 1e6 + 7;
const int M = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +7;

LL n, m, x1, x2, y1, y2;

LL exgcd(LL a, LL b, LL &x, LL &y) {
    if(!b) {
        x = 1; y = 0;
        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() {

    int T; scanf("%d", &T);
    for(int cas = 1; cas <= T; cas++) {
        LL t = -1;
        scanf("%lld%lld%lld%lld%lld%lld", &n, &m, &x1, &y1, &x2, &y2);
        n <<= 1; m <<= 1; x1 <<= 1; y1 <<= 1; x2 <<= 1; y2 <<= 1;
        printf("Case #%d:
", cas);
        LL ta = n - (x1 + x2) / 2;
        LL tb = m - (y1 + y2) / 2;

        if(x1 == x2 && y1 == y2) t = 0;
        else if(x1 == x2) t = tb;
        else if(y1 == y2) t = ta;
        else {
            LL x, y, gcd;
            gcd = exgcd(n, m, x, y);
            if((tb - ta) % gcd == 0) {

                x = (tb - ta) / gcd * x;
                x = (x % (m / gcd) + m / gcd) % (m / gcd);
                t = ta + n * x;

            }
        }

        if(t == -1) {
             puts("Collision will not happen.");
        } else {

             x1 = (x1 + t) % (2 * n);
             y1 = (y1 + t) % (2 * m);
             if(x1 > n) x1 = 2 * n - x1;
             if(y1 > m) y1 = 2 * m - y1;
             printf("%.1f %.1f
", x1 /2.0, y1 /2.0);
        }
    }
    return 0;
}


/*
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