hdu 1005 Number Sequence

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#快速幂

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本文介绍了一种使用快速幂矩阵乘法求解特定数列f(n)的方法，该数列定义为f(1)=1，f(2)=1，f(n)=(A*f(n-1)+B*f(n-2))mod7。通过构建特定矩阵并运用快速幂算法，可以在O(logn)的时间复杂度内高效计算出f(n)的值。

Problem Description

A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).

Input

The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.

Output

For each test case, print the value of f(n) on a single line.

Sample Input

1 1 3 1 2 10 0 0 0

Sample Output

2 5

Author

CHEN, Shunbao

Source

ZJCPC2004

快速幂板子题

#include <stdio.h>
#define mod 7
#define ll long long
typedef struct matrix {
    ll mat[2][2];
}matrix;
matrix unit_matrix = {1, 0, 0, 1}; //单位矩阵
matrix mul(matrix a, matrix b) {
    matrix res;
    for (int i = 0; i < 2; i++)
        for (int j = 0; j < 2; j++) {
            res.mat[i][j] = 0;
            for (int k = 0; k < 2; k++) {
                res.mat[i][j] = res.mat[i][j] + a.mat[i][k] * b.mat[k][j];
                res.mat[i][j] = res.mat[i][j] % mod;
            }
        }
    return res;
}
matrix pow_matrix(matrix a, ll n) {
    matrix res = unit_matrix;
    while (n != 0) {
        if (n & 1)
            res = mul(res, a);
        a = mul(a, a);
        n >>= 1;
    }
    return res;
}
int main() {
    ll n,a,b;
    while (scanf("%lld %lld %lld",&a,&b,&n)!=EOF)
     {
         if(a==0&&b==0&&n==0) break;
        if (n <= 2)
        {
            printf("1\n");
            continue;
        }
        matrix tmp = {a, 1, b, 0}, ans, x = {1, 1, 0, 0};
        ans = pow_matrix(tmp, n - 2);
       // printf("ans:\n");
        /*for(int i=0;i<2;i++)
        {
            for(int j=0;j<2;j++)
            {
                printf("%lld ",ans.mat[i][j]);
            }
            printf("\n");
        }*/
        //printf("\n");
        printf("%lld\n",(ans.mat[0][0]+ans.mat[1][0])%7);
    }
    return 0;
}