hdu 5318 The Goddess Of The Moon（矩阵快速幂）

题目链接：hdu 5318 The Goddess Of The Moon

将50个串处理成50*50的矩阵，注意重复串。

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

using namespace std;

const int maxn = 55;
const int mod = 1e9+7;

int N, M, A[maxn];

struct Mat {
	int s[maxn][maxn];
	Mat () {
		memset(s, 0, sizeof(s));
	}
	Mat operator * (const Mat& b) {
		Mat ret;
		for (int k = 0; k < N; k++) {
			for (int i = 0; i < N; i++) {
				for (int j = 0; j < N; j++)
					ret.s[i][j] = (ret.s[i][j] + 1LL * s[i][k] * b.s[k][j] % mod) % mod;
			}
		}
		return ret;
	}
};

void init () {
	scanf("%d%d", &N, &M);
	for (int i = 0; i < N; i++)
		scanf("%d", &A[i]);
	sort(A, A + N);
	N = unique(A, A + N) - A;
}

bool judge (int a, int b) {
	char p[15], q[15];
	sprintf(p, "%d", a);
	sprintf(q, "%d", b);

	int pp = strlen(p), qq = strlen(q);
	for (int i = 0; i < pp; i++) {
		int k = 0;
		while (i + k < pp && k < qq && p[i+k] == q[k])
			k++;
		if (i + k == pp && k > 1)
			return true;
	}
	return false;
}

Mat solve () {
	Mat ret;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			if (judge(A[i], A[j]))
				ret.s[i][j] = 1;
		}
	}

	/*
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++)
			printf("%d ", ret.s[i][j]);
		printf("\n");
	}
	*/
	return ret;
}

Mat pow_mat(Mat x, int n) {
	Mat ret;
	for (int i = 0; i < N; i++)
		ret.s[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--) {
		init();

		if (N == 0 || M == 0) {
			printf("0\n");
			continue;
		}

		Mat x = solve();
		Mat ret = pow_mat(x, M-1);

		int ans = 0;
		for (int i = 0; i < N; i++)
			for (int j = 0; j < N; j++)
				ans = (ans + ret.s[i][j]) % mod;
		printf("%d\n", ans);
	}
	return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。