OJ_快速幂

分解幂计算再加和

递推数列

• 核心：求方阵的幂
  

#include <iostream>

using namespace std;

//矩阵乘法
void MatrixMultiply(int m1[2][2],int m2[2][2],int res[2][2]){
    res[0][0] = (m1[0][0] * m2[0][0] %10000) + (m1[0][1] * m2[1][0] %10000);
    res[0][0] %= 10000;
    res[0][1] = (m1[0][0] * m2[0][1] %10000) + (m1[0][1] * m2[1][1] %10000);
    res[0][1] %= 10000;
    res[1][0] = (m1[1][0] * m2[0][0] %10000) + (m1[1][1] * m2[1][0] %10000);
    res[1][0] %= 10000;
    res[1][1] = (m1[1][0] * m2[0][1] %10000) + (m1[1][1] * m2[1][1] %10000);
    res[1][1] %= 10000;
}

//矩阵的幂
void MatrixPower(int m1[2][2],int n,int res[2][2]){
    if(n == 0){
        res[0][0] = 1;
        res[0][1] = 0;
        res[1][0] = 0;
        res[1][1] = 1;
    }else if(n%2==0){
        int temp[2][2];
        MatrixPower(m1,n/2,temp);
        MatrixMultiply(temp,temp,res);
    }else{
        int temp1[2][2];
        MatrixPower(m1,n/2,temp1);
        int temp2[2][2];
        MatrixMultiply(temp1,temp1,temp2);
        MatrixMultiply(temp2,m1,res);
    }
}

int main()
{
    int matrix[2][2];
    matrix[1][0] = 1;
    matrix[1][1] = 0;
    int a0,a1,p,q,k;
    scanf("%d%d%d%d%d",&a0,&a1,&p,&q,&k);
    matrix[0][0] = p;
    matrix[0][1] = q;
    int res[2][2];
    MatrixPower(matrix,k-1,res);
    printf("%d\n",(res[0][0]*a1%10000 + res[0][1]*a0%10000)%10000);
    return 0;
}