uva 1386 - Cellular Automaton(循环矩阵+矩阵快速幂)

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题目链接：uva 1386 - Cellular Automaton

题目大意：一个细胞自动机，有n个格子，每个格子的取值为0~m-1。给定距离d，每次操作后每个格子的值将变为到它距离不超过d的所有格子在操作之前的值之和除以m的余数。

解题思路：矩阵快速幂，因为矩阵是循环矩阵，所以记录并处理第一行即可。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 505;int N, M, D, K;struct Mat {    ll arr[maxn];};Mat ceil, x;Mat mul_mat (const Mat& a, const Mat& b) {    Mat ans;    memset(ans.arr, 0, sizeof(ans.arr));    for (int i = 0; i < N; i++) {        for (int j = 0; j < N; j++)            ans.arr[i] = (ans.arr[i] + a.arr[j] * b.arr[(i-j+N)%N]) % M;    }    return ans;}void pow_mat (int n) {    Mat ans = x;    while (n) {        if (n&1)            ans = mul_mat(ans, x);        x = mul_mat(x, x);        n >>= 1;    }    x = ans;}int main () {    while (scanf("%d%d%d%d", &N, &M, &D, &K) == 4) {        for (int i = 0; i < N; i++)            scanf("%lld", &ceil.arr[i]);        memset(x.arr, 0, sizeof(x.arr));        for (int i = -D; i <= D; i++)            x.arr[(i+N)%N] = 1;        pow_mat(K-1);        /*        for (int i = 0; i < N; i++)            printf("%lld ", x.arr[i]);        printf("\n");        */        for (int i = 0; i < N; i++) {            if (i)                printf(" ");            ll ret = 0;            for (int j = 0; j < N; j++)                ret = (ret + ceil.arr[j] * x.arr[(j-i+N)%N]) % M;            printf("%lld", ret);        }        printf("\n");    }    return 0;}

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