快速幂

1.模板

求n的m次方

#include<iostream>
#define MOD 1000000007
using namespace std;
//递归快速幂 
int quickpow(long long a,long long n){
        if(n==0) return 1;
        else if(n%2==1){
            return quickpow(a,n-1)*a%MOD;
            
        }else{
            int t=quickpow(a,n/2)%MOD;
            return t*t%MOD;
        }
}
int Uquickpow(int a,int n){
    int ans=1;
    while(n){
        if(n&1){   //n最后一位等于1 
            ans=ans*a;
        }
        a=a*a;
        n>>=1;//n右移一位 
        
    }
    return ans; 
} 
int main()
{
    
    cout<<Uquickpow(7,5);
    cout<<quickpow(7,5);


}

2.题目

洛古p1045

#include<bits/stdc++.h>
const long long mod=10000000000;
using namespace std;
const int N=1e5+10;
int l=1,la=1;
int ans[N]= {},a[N]= {},c[N]={};
using namespace std;
void cheng1(){//ans*a
    memset(c,0,sizeof(c));//临时数组c 
    for (int i = 1; i <= l; ++i)   {
        for (int j = 1; j <= la; ++j) {
            c[i+j-1] += ans[i] * a[j];
            c[i+j] += ( c[i+j-1] ) / 10;
            c[i+j-1] %= 10;
        }
    }
    int lc = l + la;
    while( c[lc] == 0 ) -- lc;
    for(int i = 1;i <= lc; ++i){
        ans[i] = c[i];
    }
    //l=lc
    if(lc>500){
        l=500;
    }else{
        l=lc;
    } 
    
    
}
void cheng2(){//a*a
    memset(c,0,sizeof(c));//临时数组c 
    for (int i = 1; i <= la; ++i)   {
        for (int j = 1; j <= la; ++j) {
            c[i+j-1] += a[i] * a[j];
            c[i+j] += ( c[i+j-1] ) / 10;
            c[i+j-1] %= 10;
        }
    }
    int lc = la + la;
    while( c[lc] == 0 ) -- lc;
    for(int i = 1;i <= lc; ++i){
        a[i] = c[i];
    }
   // la=lc;
      if(lc>500){
        la=500;
    }else{
        la=lc;
    } 
    
    
}
void qpow(int P){
    int ans=1;
    while(P){
        if(P&1){
            cheng1();//ans=ans*a
        }
        cheng2();//a=a*a
        P>>=1;
    }
}
int main()
{
    int P;
    cin>>P;
    memset( a, 0, sizeof(a));
    memset( ans, 0, sizeof(ans));
    ans[1]=1;
    a[1]=2;
    qpow(P);
//    cout<<l<<endl;
    cout<<int((P*log10(2.0)+1))<<endl;
    ans[1]--;
    int num=0;
    for(int i=500;i>=1;i--){
        cout<<ans[i];
        num++;
        if(num==50){
            cout<<endl;
            num=0;
        }
    }


}