UVA 10951 - Polynomial GCD(数论)
2017-05-01 19:27
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UVA 10951 - Polynomial GCD
题意:给定两个多项式,求多项式的gcd,要求首项次数为1,多项式中的运算都%n,而且n为素数.
思路:和gcd基本一样,仅仅只是传入的是两个多项式,因为有%n这个条件。所以计算过程能够用乘法逆去计算除法模,然后最后输出的时候每项除掉首项的次数就是答案了.
代码:
#include <stdio.h> #include <string.h> #include <vector> using namespace std; int n; vector<int> f, g; int exgcd(int a, int b, int &x, int &y) { if (!b) {x = 1; y = 0; return a;} int d = exgcd(b, a % b, y, x); y -= a / b * x; return d; } int inv(int a, int n) { int x, y; exgcd(a, n, x , y); return (x + n) % n; } vector<int> pmod(vector<int> f, vector<int> g) { int fz = f.size(), gz = g.size(); for (int i = 0; i < fz; i++) { int k = fz - i - gz; if (k < 0) break; int a = f[i] * inv(g[0], n) % n; for (int j = 0; j < gz; j++) { int now = i + j; f[now] = ((f[now] - a * g[j] % n) % n + n) % n; } } vector<int> ans; int p = -1; for (int i = 0; i < fz; i++) if (f[i] != 0) {p = i; break;} if (p >= 0) for (int i = p; i < fz; i++) ans.push_back(f[i]); return ans; } vector<int> gcd(vector<int> f, vector<int> g) { if (g.size() == 0) return f; return gcd(g, pmod(f, g)); } int main() { int cas = 0; while (~scanf("%d", &n) && n) { f.clear(); g.clear(); int a, k; scanf("%d", &a); for (int i = 0; i <= a; i++) { scanf("%d", &k); f.push_back(k); } scanf("%d", &a); for (int i = 0; i <= a; i++) { scanf("%d", &k); g.push_back(k); } vector<int> ans = gcd(f, g); int tmp = inv(ans[0], n), ansz = ans.size();; printf("Case %d: %d", ++cas, ansz - 1); for (int i = 0; i < ansz; i++) { printf(" %d", ans[i] * tmp % n); } printf("\n"); } return 0; }
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