题目链接：uva 11149 - Power of Matrix

题目大意：给定一个矩阵，求∑ikAi

解题思路：因为k比较大，所以即使用快速幂的话复杂度还是有点高，利用矩阵倍增的方法∑ikAi=(1+Ak/2)?∑ik/2Ai

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

using namespace std;
const int maxn = 50;
const int MOD = 10;

struct Mat {
    int r, c, arr[maxn][maxn];
    Mat () { memset(arr, 0, sizeof(arr)); }
    Mat (int r = 0, int c = 0) { set(r, c); }
    void set (int r, int c) {
        this->r = r;
        this->c = c;
        memset(arr, 0, sizeof(arr));
    }

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

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

int N, K;

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

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

Mat solve (Mat x, int n) {
    if (n == 1)
        return x;

    Mat ret(N, N);
    for (int i = 0; i < N; i++)
        ret.arr[i][i] = 1;

    if (n == 0)
        return ret;
    ret = (ret + pow_mat(x, n>>1)) * solve(x, n>>1);

    if (n&1)
        ret = ret + pow_mat(x, n);
    return ret;
}

void put (Mat u) {
    for (int i = 0; i < u.r; i++) {
        printf("%d", u.arr[i][0]);
        for (int j = 1; j < u.c; j++)
            printf(" %d", u.arr[i][j]);
        printf("\n");
    }
}

int main () {
    while (scanf("%d%d", &N, &K) == 2 && N) {
        Mat ans(N, N);
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                scanf("%d", &ans.arr[i][j]);
                ans.arr[i][j] %= MOD;
            }
        }

        ans = solve(ans, K);
        put(ans);
        printf("\n");
    }
    return 0;
}