uva 10689 - Yet another Number Sequence(矩阵快速幂)

题目链接：uva 10689 - Yet another Number Sequence

题目大意：给定斐波那契数列头两项，求的n项取模10m

解题思路：矩阵快速幂。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 5;
const int M[maxn+5] = {1, 10, 100, 1000, 10000};

int MOD;

struct Mat {
    int arr[maxn][maxn];
    Mat () {
        memset(arr, 0, sizeof(arr));
    }

    Mat operator * (const Mat& u) {
        Mat ret;
        for (int k = 0; k < 2; k++) {
            for (int i = 0; i < 2; i++)
                for (int j = 0; j < 2; j++)
                    ret.arr[i][j] = (ret.arr[i][j] + arr[i][k] * u.arr[k][j]) % MOD;
        }
        return ret;
    }
};

Mat pow_mat(Mat x, int n) {
    Mat ret;
    for (int i = 0; i < 2; i++)
        ret.arr[i][i] = 1;

    while (n) {
        if (n&1) 
            ret = ret * x;
        x = x * x;
        n >>= 1;
    }
    return ret;
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        int a, b, n, m;
        scanf("%d%d%d%d", &a, &b, &n, &m);
        MOD = M[m];

        Mat x;
        for (int i = 0; i < 2; i++)
            for (int j = 0; j < 2; j++)
                x.arr[i][j] = 1;
        x.arr[0][0] = 0;

        if (n >= 1) {
        Mat ans = pow_mat(x, n-1);
        printf("%d\n", (ans.arr[1][0] * a % MOD + ans.arr[1][1] * b % MOD) % MOD);
        } else
            printf("%d\n", a);
    }
    return 0;
}