uva1386(矩陣快速冪)

題意：

一個細胞自動機，有n個格子，每個格子的取值爲0~m-1。給定距離d，每次操作後每個格子的值將變爲到它距離不超過d的所有格子在操作之前的值之和除以m的餘數。

思路：

矩陣快速冪

#include <cstdio>
#include <cstring>
#define ll long long
const int N = 505;

int n, M, D, K;

struct Mat {
    ll arr[N];
};
Mat ceil, x;

Mat mul (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 (int n) {
    Mat ans = x;

    while (n) {
        if (n&1)
            ans = mul(ans, x);
        x = mul(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(K-1);
        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;
}