uva 10951 - Polynomial GCD(欧几里得)

题目链接：uva 10951 - Polynomial GCD

题目大意：给出n和两个多项式，求两个多项式在所有操作均模n的情况下最大公约数是多少。

解题思路：欧几里得算法，就是为多项式这个数据类型重载取模运算符，需要注意的是在多项式除多项的过程中，为了保证各项系数为整数，需要将整个多项式的系数整体扩大至一定倍数，碰到先除后模的时候要用逆元。

#include <cstdio>
#include <cstring>

const int maxn = 105;
int M;

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

int inv (int a, int b) {
    int d, x, y;
    gcd(a, b, d, x, y);
    return (x + b) % b;
}

struct state {
    int d;
    int x[maxn];

    state() {
        d = 0;
        memset(x, 0, sizeof(x));
    }

    void get () {
        scanf("%d", &d);
        for (int i = d; i >= 0; i--)
            scanf("%d", &x[i]);
    }

    void put () {
        printf(" %d", d);
        for (int i = d; i >= 0; i--)
            printf(" %d", x[i]);
    }

    state operator * (const int& a) {
        state ans = *this;
        for (int i = 0; i <= d; i++)
            ans.x[i] = ans.x[i] * a % M;
        ans.del();
        return ans;
    }

    state operator - (const state& a) {

        state ans = *this;
        for (int i = 0; i <= a.d; i++)
            ans.x[d-i] = (ans.x[d-i] - a.x[a.d-i] + M) % M;
        ans.del();
        return ans;
    }

    void del () {
        for (int i = d; i >= 0; i--)
            x[d] = (x[d] + M) % M;

        while (d > 0 && x[d] == 0)
            d--;
    }

    bool empty() {
        return d <= 0 && x[0] == 0;
    }
};

state mod (state a, state b) {
    for (int i = a.d; i >= b.d; i--) {
        int p = a.x[i], q = b.x[b.d];

        state c = b * inv(q, M) * p;;
        a = a - c;
        /*
        a.put();
        printf("\n");
        */
    }
    return a;
}

state sgcd (state a, state b) {
    return b.empty() ? a : sgcd(b, mod(a, b));
}

int main () {
    int cas = 1;
    while (scanf("%d", &M) == 1 && M) {
        state a, b;
        a.get();
        b.get();

        state c = sgcd(a, b);

        printf("Case %d:", cas++);
        c = c * inv(c.x[c.d], M);
        c.put();
        printf("\n");
    }
    return 0;
}