高精度快速幂 | 洛谷 | P1045

https://www.luogu.org/problemnew/show/P1045

---

高精度板子:
快速幂:

ll quickPower(ll a,ll b){
    ll ans=1,base=a;
    while(b>0){
        if(b&1)
            ans*=base;
        base*=base;
        b>>=1;
    }
    return ans;
}

还涉及到取对数的问题

#include <bits/stdc++.h>
#define FF(a,b) for(int a=0;a<b;a++)
#define F(a,b) for(int a=1;a<=b;a++)
#define LEN 510000
#define INF 1000000
#define bug(x) cout<<#x<<"="<<x<<endl;

using namespace std;

int P;
int L=500;

struct hp{
    int len;
    int *s;
    hp(){
        s=new int [LEN];
        memset(s,0,4*LEN);
        len=1;

    }
    hp(char* ch){
        len=strlen(ch);
        s=new int [LEN];
        memset(s,0,4*LEN);
        for(int i=1;i<=len;i++)
            s[i]=ch[len-i]-48;
    }
    void print(){
        int i;
        for(i=500;i>=1;i--){
            if(i%50==0) puts("");
            printf("%d",s[i]);
        }

    }
};

hp multiplyh(const hp& a,const hp& b){
    int i,j,len;
    hp c;
    for(i=1;i<=a.len;i++){
        for(j=1;j<=b.len;j++){
            c.s[i+j-1]+=a.s[i]*b.s[j];
            c.s[i+j]+=c.s[i+j-1]/10;
            c.s[i+j-1]%=10;
        }
    }
    len=a.len+b.len+1;
    while(len>1 && c.s[len]==0)len--;
    c.len=len;
    if(c.len>=L){//截取最后500项
        c.len=500;
    }
    return c;
}

int main()
{
    freopen("./in","r",stdin);
    scanf("%d",&P);
    printf("%d",int( log10(2)*P + 1) );
    hp ans("1");
    hp base("2");
    while(P>0){
        if(P&1)
            ans=multiplyh(ans,base);
        base= multiplyh(base,base);
        P>>=1;
    }
    ans.s[1]-=1;
    ans.print();
    return 0;
}