题目链接：uva 10951 - Polynomial GCD
题目大意：给出n和两个多项式，求两个多项式在全部操作均模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，须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，须要将整个多项式的系数总体扩大至一定倍数，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;
}