hiho 1143 矩阵快速幂 求递推式

// m^n % k
int quickpow(int m,int n,int k)
{
    int b = 1;
    while (n > 0)
    {
          if (n & 1)
             b = (b*m)%k;
          n = n >> 1 ;
          m = (m*m)%k;
    }
    return b;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#define MOD 19999997
using namespace std;
struct Matrix{
    long long m[2][2];   //一定要long long   
};

Matrix mult(Matrix a,Matrix b)
{
    Matrix t;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
         t.m[i][j]=(a.m[i][0]*b.m[0][j])%MOD+(a.m[i][1]*b.m[1][j])%MOD;  //long long 不会爆

    return t;
}

int m_fastpow(int n)
{
    Matrix unit={1,0,0,1};       //单位矩阵
    Matrix x={0,1,1,1};          //递推式
    while(n>0)
    {
        if(n&1)
            unit=mult(x,unit);
        x=mult(x,x);
        n=n>>1;
    }
    return unit.m[1][1]%MOD;
}

int main()
{
    int n;
    scanf("%d",&n);
    printf("%d\n",m_fastpow(n));
    return 0;
}