bzoj 1407 扩展欧几里德

思路：枚举洞穴个数，用扩展欧几里德暴力判断没两个人的周期。

#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 = 15 + 7;
const int M = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +7;

LL n, c[N], p[N], l[N];

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;
    }
}

bool check(LL m) {
    for(int i = 1; i < n; i++) {
        for(int j = i + 1; j <= n; j++) {
            LL a = p[i] - p[j], b = m, val = c[j] - c[i];
            LL x, y;
            LL gcd = exgcd(a, b, x, y);

            if(val % gcd) continue;

            x *= val / gcd;
            x %= (b / gcd);
            if(x < 0) x += abs(b / gcd);
            if(x <= min(l[i], l[j])) return false;
        }
    }
    return true;
}
int main() {
    scanf("%lld", &n);
    LL mx = 1;
    for(int i = 1; i <= n; i++) {
        scanf("%lld%lld%lld", &c[i], &p[i], &l[i]);
        mx = max(mx, c[i]);
    }

    for(int i = mx; ; i++) {
        if(check(i)) {
            printf("%d\n", i);
            break;
        }
    }

    return 0;
}


/*
*/