UVA - 10951 Polynomial GCD 【最大公共多项式】

Disctiption
Given a decimal integer number you will have to find out how many trailing zeros will be there in its
factorial in a given number system and also you will have to find how many digits will its factorial have
in a given number system? You can assume that for a b based number system there are b different
symbols to denote values ranging from 0 . . . b − 1.

Input
There will be several lines of input. Each line makes a block. Each line will contain a decimal number
N (a 20bit unsigned number) and a decimal number B (1 < B ≤ 800), which is the base of the number
system you have to consider. As for example 5! = 120 (in decimal) but it is 78 in hexadecimal number
system. So in Hexadecimal 5! has no trailing zeros.

Output
For each line of input output in a single line how many trailing zeros will the factorial of that number
have in the given number system and also how many digits will the factorial of that number have in
that given number system. Separate these two numbers with a single space. You can be sure that the
number of trailing zeros or the number of digits will not be greater than 231 − 1.

Sample Input
2 10
5 16
5 10

Sample Output
0 1
0 2
1 3

题意
给定两个多项式，求塔恩两个的最大公共多项式。

思路
跟求最大公因子思路相似。
套了个板子。
求最大公共多项式

AC代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

vector<int>G[100010],ans;
int n;
int a, b;

ll qmi(ll m, ll k, ll mod)
{
    ll res = 1 % mod, t = m;
    while (k)
    {
        if (k&1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}

vector<int> poly_gcd(vector<int> a,vector<int> b)
{
    if (b.size() == 0)
        return a;
    int t = a.size() - b.size();
    vector<int> c;
    for (int i=0; i<=t; i++)
    {
        int tmp=a[i]*qmi(b[0],n-2,n)%n;
        for(ll j=0; j<b.size(); j++)
            a[i+j]=(a[i+j]-tmp*b[j]%n+n)%n;
    }
    int p=-1;
    for(int i=0; i<a.size(); i++)
    {
        if(a[i]!=0)
        {
            p=i;
            break;
        }
    }
    if(p>=0)
    {
        for(int i=p; i<a.size(); i++)
            c.push_back(a[i]);
    }
    return poly_gcd(b,c);
}

int main()
{
    int cas=1;
    while(scanf("%d", &n)!=EOF)
    {
        if(n==0)
            return 0;
        G[0].clear();
        G[1].clear();
        for(int i=0; i<2; i++)
        {
            scanf("%d", &a);
            for (int j=0; j<=a; j++)
            {
                scanf("%d", &b);
                G[i].push_back(b);
            }
        }
        ans=poly_gcd(G[0], G[1]);
        printf("Case %d: %d", cas, ans.size()-1);
        int tmp=qmi(ans[0], n-2,n);
        for(int i=0; i<ans.size(); i++)
        {
            ans[i]=ans[i]*tmp%n;
            printf(" %d", ans[i]);
        }
        printf("\n");
        cas++;
    }
    return 0;
}