BZOJ 1951 SDOI2010古代猪文

数论大杂烩，留坑代填，先放代码：

/**************************************************************
    Problem: 1951
    User: JBLee
    Language: C++
    Result: Accepted
    Time:152 ms
    Memory:2856 kb
****************************************************************/
 
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int maxn=1e5+7;
const int mod=999911658;
int n,g;
int sum;
int m[5]={0,2,3,4679,35617};
int jc[maxn];
int x,y;
int china[maxn];
int ksm(int x,int n,int mo){
    int base=1;
    while(n){
        if(n&1) base=base*x%mo;
        x=x*x%mo;
        n>>=1; 
    }
    return base;
}
int exgcd(int a,int &x,int b,int &y){
    if(!b){
        x=1;y=0;
        return a;
    }
    int gcd=exgcd(b,x,a%b,y);
    int tmp=x;x=y;y=tmp-(a/b)*y;
    return gcd;
}
int inv(int a,int mo){
    exgcd(a,x,mo,y);
    return (x%mo+mo)%mo;
}
void work(int p){
    jc[0]=1,jc[1]=1;
    for(int i=2;i<=p;i++) jc[i]=jc[i-1]*i%p;
}
int comb(int n,int m,int p){
    if(m>n) return 0;
    return jc[n]*inv(jc[n-m],p)%p*inv(jc[m],p)%p;
}
int lucas(int n,int m,int p){
    if(!m) return 1;
    else return comb(n%p,m%p,p)*lucas(n/p,m/p,p)%p;
}
int solve(int p){
    work(p);
    int ans=0;
    for(int i=1;i*i<=n;i++){
        if(n%i==0){
            ans=(ans+lucas(n,i,p))%p;
            if(n/i!=i) ans=(ans+lucas(n,n/i,p))%p;
        }
    }
    return ans;
}
int china_mod(){
    for(int i=1;i<=4;i++) china[i]=solve(m[i]);
    int lcm=1,ans=0,mi;
    for(int i=1;i<=4;i++) lcm*=m[i];
    for(int i=1;i<=4;i++){
        mi=lcm/m[i];
        exgcd(mi,x,m[i],y);
        ans=((ans+mi*x*china[i])%lcm+lcm)%lcm;
    }
    return ans;
} 
signed main(){
    scanf("%lld%lld",&n,&g);
    if(g%(mod+1)==0){
        printf("0\n");
        return 0;
    }
    printf("%lld\n",ksm(g,china_mod(),mod+1));
    return 0;
} 

标签：return,int,mo,x%,SDOI2010,base,1951,猪文,jc
来源： https://www.cnblogs.com/LJB666/p/11779219.html