uva 11551 - Experienced Endeavour(矩阵快速幂)

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题目链接：uva 11551 - Experienced Endeavour

题目大意：给定一个序列的变换，求变换r次后各项的值。

解题思路：矩阵快速幂，不解释。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 55;const int MOD = 1000;struct Mat {    int r, c, arr[maxn][maxn];    Mat (int r = 0, int c = 0) {        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 pow_mat (Mat ret, Mat x, int n) {    while (n) {        if (n&1)            ret = x * ret;        x = x * x;        n >>= 1;    }    return ret;}int main () {    int cas, N, K;    scanf("%d", &cas);    while (cas--) {        scanf("%d%d", &N, &K);        Mat a(N, 1);        for (int i = 0; i < N; i++) {            scanf("%d", &a.arr[i][0]);            a.arr[i][0] %= MOD;        }        int x, pos;        Mat b(N, N);        for (int i = 0; i < N; i++) {            scanf("%d", &x);            for (int j = 0; j < x; j++) {                scanf("%d", &pos);                b.arr[i][pos] = 1;            }        }        a = pow_mat(a, b, K);        printf("%d", a.arr[0][0]);        for (int i = 1; i < N; i++)            printf(" %d", a.arr[i][0]);        printf("\n");    }    return 0;}

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