[BZOJ3283]运算器

description

BZOJ

solution

(exbsgs+exlucas+excrt)模板题。
写到手炸...

#include<bits/stdc++.h>
#define FL "3283"
using namespace std;
typedef long long ll;
const int N=1e5+10;
const int mod1=1e6+3;
const int mod2=998244353;
inline int read(){
  int data=0,w=1;char ch=getchar();
  while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
  if(ch=='-')w=-1,ch=getchar();
  while(ch<='9'&&ch>='0')data=data*10+ch-48,ch=getchar();
  return data*w;
}
inline void file(){
  freopen(FL".in","r",stdin);
  freopen(FL".out","w",stdout);
}

namespace math{
  inline int gcd(int a,int b){return b?gcd(b,a%b):a;}
  inline void exgcd(int a,int b,int &x,int &y,int &d){
    if(!b){d=a;x=1;y=0;return;}exgcd(b,a%b,y,x,d);y-=a/b*x;
  }
  inline int inv(int a,int p){
    int x,y,d;exgcd(a,p,x,y,d);return (x%p+p)%p;
  }
  inline int poww(int a,int b,int mod){
    int res=1;
    for(;b;b>>=1,a=1ll*a*a%mod)
      if(b&1)res=1ll*res*a%mod;
    return res;
  }
  inline int excrt(int n,int *a,int *p){
    for(int i=2,x,y,d,L;i<=n;i++){
      exgcd(p[1],p[i],x,y,d);L=p[1]/d*p[i];
      x=(1ll*x*(a[i]-a[1])%p[i]+p[i])%p[i];
      a[1]=(1ll*x*p[1]%L+a[1])%L;p[1]=L;
    }
    return a[1];
  }
}
using math::inv;
using math::poww;
using math::excrt;
using math::gcd;

namespace HASH{
  int head[mod1],nxt[N],to[N],val[N],cnt;
  inline void init(){memset(head,-1,sizeof(head));while(cnt)head[cnt--]=-1;}
  inline void insert(int a,int b){
    to[++cnt]=b;val[cnt]=a%mod2;
    nxt[cnt]=head[a%mod1];head[a%mod1]=cnt;
  }
  inline int find(int a){
    int r=-1;
    for(int i=head[a%mod1];i!=-1;i=nxt[i])
      if(a%mod2==val[i]){
	if(r==-1)r=to[i];
	else r=min(r,to[i]);
      }
    return r;
  }
}
using HASH::init;
using HASH::insert;
using HASH::find;

namespace Exbsgs{
  inline int logmod(int a,int b,int p){
    int q=sqrt(p)+1;init();
    for(int i=0,r=1%p;i<=q;i++,r=1ll*r*a%p)insert(r,i);
    for(int i=0,v=inv(poww(a,q,p),p),r;i<=q;i++,b=1ll*b*v%p)
      if(r=find(b),r!=-1)return i*q+r;
    return -1;
  }
  inline int exbsgs(int a,int b,int p){
    int cnt=0,r=1,d;
    while(gcd(a,p)!=1){
      if(r==b)return cnt;
      d=gcd(a,p);
      if(b%d)return -1;
      b/=d;p/=d;r=1ll*r*(a/d)%p;cnt++;
    }
    return logmod(a,1ll*b*inv(r,p)%p,p)+cnt;
  }
}
using Exbsgs::exbsgs;

namespace Exlucas{
  int p[20],pk[20],fac[20][N],r[20],tot;
  inline void fact(int x){
    tot=0;
    for(int i=2;i*i<=x;i++)
      if(x%i==0){
	p[++tot]=i;pk[tot]=1;
	while(x%i==0)pk[tot]*=i,x/=i;
      }
    if(x>1)tot++,p[tot]=pk[tot]=x;
    for(int i=1;i<=tot;i++){
      fac[i][0]=1;
      for(int j=1;j<=pk[i];j++)
	fac[i][j]=1ll*fac[i][j-1]*(j%p[i]?j:1)%pk[i];
    }
  }
  inline int mul(int i,int n){
    int res=poww(fac[i][pk[i]],n/pk[i],pk[i]);
    for(int j=n/pk[i]*pk[i]+1;j<=n;j++)
      res=1ll*res*(j%p[i]?j:1)%pk[i];
    return 1ll*res*(n/p[i]?mul(i,n/p[i]):1)%pk[i];
  }
  inline int exlucas(int n,int m,int mod){
    if(n<m)return 0;fact(mod);
    for(int i=1;i<=tot;i++){
      int a=mul(i,n),b=mul(i,m),c=mul(i,n-m),k=0;
      for(int j=n;j;j/=p[i])k+=j/p[i];
      for(int j=m;j;j/=p[i])k-=j/p[i];
      for(int j=n-m;j;j/=p[i])k-=j/p[i];
      r[i]=1ll*a*inv(b,pk[i])%pk[i]*inv(c,pk[i])%pk[i]*poww(p[i],k,pk[i])%pk[i];
    }
    return excrt(tot,r,pk);
  }
}
using Exlucas::exlucas;

int main()
{
  init();
  int T=read();
  while(T--){
    int opt=read(),y=read(),z=read(),p=read(),r;
    if(opt==1)printf("%d
",poww(y,z,p));
    if(opt==2){
      r=exbsgs(y,z,p);
      r==-1?puts("Math Error"):printf("%d
",r);
    }
    if(opt==3)printf("%d
",exlucas(z,y,p));
  }
  return 0;
}