6492. 【GDOI2020模拟03.04】多项式

题目描述

假的，n<=10^5

题解

（本题中倍数可以为负）

满足条件的p要么是|Ai|（|Ai|≠0）的gcd的约数，要么原式中存在\(\prod_{i=0}^{p-1}{(x-i)}\)

神奇结论：\(\prod_{i=0}^{p-1}{(x-i)}=x^p-x\;(mod\;p) \; p \in prime\)

证明： https://www.cnblogs.com/Dup4/p/10750749.html

大概是右边=x(x^(p-1)-1)，根据费马小定理等于0，和左边的点值完全一致

随便枚举一下p，多项式除法O(n)判断即可

据说会卡常？

code

#include <bits/stdc++.h>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define abs(x) ((x)>0?(x):-(x))
#define ll long long
#define file
using namespace std;

ll a[100001],b[100001],s;
int p[100001],ans[200001],n,i,j,k,l,tot,len,S;
bool f[100001];

ll gcd(ll n,ll m)
{
    ll r=n%m;
    
    while (r)
    n=m,m=r,r=n%m;
    
    return m;
}

void init()
{
    int i,j,k,l;
    
    fo(i,2,n)
    {
        if (!f[i])
        p[++len]=i;
        
        fo(j,1,len)
        if (1ll*i*p[j]<=n)
        {
            f[i*p[j]]=1;
            
            if (!(i%p[j]))
            break;
        }
        else
        break;
    }
}

void pd(int p)
{
    int i;
    
    fd(i,n,p)
    {
        b[i-p+1]=(b[i-p+1]+b[i])%p;
        b[i]=0;
    }
    fo(i,0,n) if (b[i]) return;
    
    ans[++tot]=p;
}

int main()
{
    freopen("poly.in","r",stdin);
    #ifdef file
    freopen("poly.out","w",stdout);
    #endif
    
    scanf("%d",&n);
    fd(i,n,0)
    {
        scanf("%lld",&a[i]);
        
        if (a[i])
        s=(s)?gcd(s,abs(a[i])):abs(a[i]);
    }
    
    init();
    
    fo(i,1,len)
    if (s%p[i] && !(a[0]%p[i]))
    {
        fo(j,0,n) b[j]=a[j]%p[i];
        pd(p[i]);
    }
    
    S=floor(sqrt(s));
    fo(i,2,S)
    if (!(s%i))
    {
        ans[++tot]=i;
        while (!(s%i))
        s/=i;
    }
    if (s>1) ans[++tot]=s;
    
    if (tot)
    sort(ans+1,ans+tot+1);
    
    fo(i,1,tot)
    printf("%d\n",ans[i]);
    
    fclose(stdin);
    fclose(stdout);
    
    return 0;
}