BZOJ4891 TJOI2017龙舟（Polllard-Rho）

　　对给定模数分解质因数后约分即可。依然常数巨大过不了。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 10010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
ll gcd(ll n,ll m){return m==0?n:gcd(m,n%m);}
ll read()
{
    ll x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,q,cnt,p[5000010],prime[500010],tot[100],t;
ll g[100],a[25][N],b[N],c[N],d[N];
bool flag[5000010];
ll ksc(ll a,ll b,ll p)
{
    ll t=a*b-(ll)((long double)a*b/p+0.5)*p;
    return t<0?t+p:t;
}
ll ksm(ll a,ll k,ll p)
{
    ll s=1;
    for (;k;k>>=1,a=ksc(a,a,p)) if (k&1) s=ksc(s,a,p);
    return s;
}
void exgcd(ll &x,ll &y,ll a,ll b)
{
    if (b==0)
    {
        x=1,y=0;
        return;
    }
    exgcd(x,y,b,a%b);
    ll t=x;x=y;y=t-a/b*x;
}
ll inv(ll a,ll p)
{
    ll x,y;exgcd(x,y,a,p);
    x=(x%p+p)%p;
    return x; 
}
bool check(int k,ll n)
{
    if (ksm(k,n-1,n)!=1) return 0;
    ll p=n-1;
    while (!(p&1))
    {
        p>>=1;ll x=ksm(k,p,n);
        if (x==n-1) return 1;
        if (x!=1) return 0;
    }
    return 1;
}
bool Miller_Rabin(ll n)
{
    if (n<=5000000) return !flag[n];
    return check(2,n)&&check(3,n)&&check(5,n)&&check(7,n)&&check(61,n)&&check(24251,n);
}
ll f(ll x,ll p,int c){return (ksc(x,x,p)+c)%p;}
void getfactor(ll n)
{
    if (n<=5000000)
    {
        while (n>1) g[++cnt]=p[n],n/=p[n];
        return;
    }
    if (Miller_Rabin(n)) {g[++cnt]=n;return;}
    while (1)
    {
        int c=rand()%(n-1)+1;
        ll x=rand()%n,y=x;
        do
        {
            ll z=gcd(abs(x-y),n);
            if (z>1&&z<n) {getfactor(z),getfactor(n/z);return;}
            x=f(x,n,c),y=f(f(y,n,c),n,c);
        }while (x!=y);
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4891.in","r",stdin);
    freopen("bzoj4891.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read(),q=read();
    flag[1]=1;
    for (int i=2;i<=5000000;i++)
    {
        if (!flag[i]) prime[++t]=i,p[i]=i;
        for (int j=1;j<=t&&prime[j]*i<=5000000;j++)
        {
            flag[prime[j]*i]=1;
            p[prime[j]*i]=prime[j];
            if (i%prime[j]==0) break;
        }
    }
    for (int i=1;i<=m;i++) b[i]=read();
    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
        a[i][j]=read();
    while (q--)
    {
        int x=read();ll y=read(),ans=1;cnt=0;getfactor(y);
        sort(g+1,g+cnt+1);cnt=unique(g+1,g+cnt+1)-g-1;memset(tot,0,sizeof(tot));
        memcpy(c,b,sizeof(c));memcpy(d,a[x],sizeof(d));
        for (int i=1;i<=m;i++)
        {
            for (int j=1;j<=cnt;j++)
            while (c[i]%g[j]==0) tot[j]++,c[i]/=g[j];
            for (int j=1;j<=cnt;j++)
            while (d[i]%g[j]==0) tot[j]--,d[i]/=g[j];
            ans=ksc(ksc(ans,c[i],y),inv(d[i],y),y);
        }
        for (int i=1;i<=cnt;i++)
        if (tot[i]<0) {ans=-1;break;}
        else ans=ksc(ans,ksm(g[i],tot[i],y),y);
        cout<<ans<<endl;
    }
    return 0;
}