51 Nod 1013 3的幂的和 矩阵链乘法||逆元+快速幂

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int mod=1000000007;
int read(){
    int ans=0,f=1,c=getchar();
    while(c<‘0‘||c>‘9‘){if(c==‘-‘) f=-1; c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){ans=ans*10+(c-‘0‘); c=getchar();}
    return ans*f;
}
int n;
int qmod(long long a,int b){
    long long ans=1; a%=mod;
    while(b){
        if(b&1) ans=ans*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return ans;
}
int main()
{
    n=read()+1;
    printf("%lld\n",(qmod(3,n)-1)*(long long)500000004%mod);
    return 0;
}

#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
using namespace std;
const int mod=1000000007;
int read(){
    int ans=0,f=1,c=getchar();
    while(c<‘0‘||c>‘9‘){if(c==‘-‘) f=-1; c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){ans=ans*10+(c-‘0‘); c=getchar();}
    return ans*f;
}
typedef LL mat[2][2];
int n;
void quickmod(mat a,mat b){
    mat c={0};
    for(int i=0;i<2;i++)
     for(int k=0;k<2;k++)
      for(int j=0;j<2;j++)
       c[i][j]=(c[i][j]+a[i][k]*b[k][j])%mod;
    for(int i=0;i<2;i++)
     for(int j=0;j<2;j++)
      a[i][j]=c[i][j];
}
void fastpow(int n){
    mat a={1,0,0,1},b={3,1,0,1};
    while(n){
        if(n&1) quickmod(a,b);
        n>>=1; 
        quickmod(b,b);
    }
    printf("%d\n",(a[0][0]+a[0][1])%mod);
}
int main()
{
    n=read(); fastpow(n);
    return 0;
}