P1951: [Sdoi2010]古代猪文

 1 const maxn=999911659;
 2 var n,g,i,j,ans:longint;
 3 tem:int64;
 4 function pow(g,x:longint):longint;
 5 var
 6 now,tem,i,t:int64;
 7 begin
 8   tem:=1; now:=g;
 9   if g=1 then exit(1);
10   while now<maxn do
11     begin
12       now:=now*now;
13       tem:=tem*2;
14     end;
15   now:=now mod maxn;
16   if (now=0) and (x div tem>=1) then exit(0);
17   if x div tem>=1 then t:=pow(now,x div tem)
18     else t:=1;
19   now:=g; x:=x-tem*(x div tem);
20   while x<>0 do
21     begin
22       if (x mod 2)=1 then begin
23         t:=(t*now) mod maxn;
24       end;
25       now:=(now*now) mod maxn;
26       x:=x div 2;
27     end;
28   exit(t mod maxn);
29 end;
30 function c(x:longint):longint;
31 var i:longint;
32 tem:int64;
33 begin
34   tem:=g;
35   for i:=1 to x do
36     tem:=pow(tem,n-i+1) mod maxn;
37   for i:=1 to x do tem:=trunc(exp(ln(tem)/i));
38   exit(tem);
39 end;
40 begin
41   readln(n,g);
42   tem:=1;
43   for i:=1 to trunc(sqrt(n)) do
44     if n mod i=0 then
45       begin
46         if i*i<>n then tem:=tem*((c(i) mod maxn)*(c(n div i) mod maxn) mod maxn) mod maxn
47           else tem:=(tem*c(i)) mod maxn;
48       end;
49   writeln(tem mod maxn);
50 end.

#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstdio>
using namespace std;
#define maxn 35620
typedef long long LL;
#define MOD 999911659
#define M 999911658
LL w[4]={2,3,4679,35617};
LL a[4];
LL fac[4][maxn];
LL Pow(LL a,LL b,LL mod)//快速幂
{
    LL ans=1;
    while(b)
    {
        if(b&1)  ans=(ans*a)%mod;
        a=(a*a)%mod;
        b >>= 1;
    }
    return ans;
}
void init()//预处理阶乘
{
    for(int i=0;i<4;i++)
    {
        fac[i][0]=1;
        for(int j=1;j<=w[i];j++)
        {
            fac[i][j]=(fac[i][j-1]*j)%w[i];
        }
    }
}
/********************************
*  组合数取模用费马小定理
*********************************/
LL C(LL n,LL m,int x)//组合数取模
{
    if(n < m) return 0;
    return (fac[x][n] * Pow((fac[x][n-m]*fac[x][m]),w[x]-2,w[x]))%w[x];
}
/*******************************
*    lucas 处理大组合数取模
********************************/
LL Lucas(LL n,LL m,int x)//lucas定理
{
    if(m==0) return 1;
    return (Lucas(n/w[x],m/w[x],x)*C(n%w[x],m%w[x],x))%w[x];
}
/****************************
*   扩展欧几里得求乘法逆元
*****************************/
LL exgcd(LL a,LL b,LL &x,LL &y)//乘法逆元
{
    if(!b){x = 1;y = 0;return a;}
    
    LL ans = exgcd(b,a%b,x,y);
    
    LL t = x;x = y;y = t - a/b*y;
    
    return ans;
}
/***************************************************************************************
*      中国剩余定理：
*       x = b1 % m1 
*       x = b2 % m2
*       x = b3 % m3
*       .
*       gcd(m1,m2,m3,...) = 1;
*       M = m1 * m2 * m3 *.....;
*       M1 = m2 * m3 * ...., M2 = m1 * m3 * ...., M3 = m1 * m2 * m4 *....., ......;
*       M1 * M(-1) = 1 % M ,M2 * M2(-1) = 1 % M;
*       res = (M1(-1)*b1 + M2(-1)*b2+.....)%M;res即为所求值
*       注：如果取模的值相同：都是m1  那么 bn的值可以相加计算；
*       略屌。。。。。。。。。。。。。。
*
*****************************************************************************************/
LL CRT()//计算组合数和取模之后的值
{
    LL i,d,x0,y0,ans=0;
    for(i = 0;i < 4;i++)//中国剩余定理
    {
        d=M/w[i];
        exgcd(d,w[i],x0,y0);
        ans=(ans+d*x0*a[i])%M;
    }
    if(ans <= 0) ans += M;
    return ans;
}
int main()
{
    init();
    LL g,n;
    while(cin>>n>>g)
    {
        memset(a,0,sizeof(a));
        for(int i=1;i*i<=n;i++)
        {
            if(n%i==0)
            {
                LL tmp=n/i;
                for(int j=0;j<4;j++)
                {
                    if(tmp!=i) a[j]=(a[j]+Lucas(n,i,j))%w[j];
                    a[j]=(a[j]+Lucas(n,tmp,j))%w[j];
                }
            }
        }
        cout<<Pow(g%MOD,CRT(),MOD)<<endl;
    }
    return 0;
}