uva 10743 - Blocks on Blocks(矩阵快速幂)

题目链接：uva 10743 - Blocks on Blocks

题目大意：问说n联骨牌有多少种，旋转镜像后相同不算同一组，一行的格子必须连续，如果答案大于10000，输出后四位。

解题思路：想了一下午的递推式，实在受不了，把推出的序列在网上搜了一下，有公式ai=5?ai?1?7?ai?2+4?ai?3
(i≥5)

PS:哪位神人知道怎么推出来的请留言我，大恩不言谢~

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

using namespace std;
const int maxn = 3;
const int mod = 10000;
int sign[maxn][maxn] = { {0, 1, 0}, {0, 0, 1}, {4, -7, 5} };

struct MAT {
    int s[maxn][maxn];

    MAT () {}

    void set () {
        for (int i = 0; i < maxn; i++)
            for (int j = 0; j < maxn; j++)
                s[i][j] = sign[i][j];
    }

    MAT operator * (const MAT& a) {
        MAT ans;
        for (int i = 0; i < maxn; i++) {
            for (int j = 0; j < maxn; j++) {
                ans.s[i][j] = 0;
                for (int k = 0; k < maxn; k++)
                    ans.s[i][j] = (ans.s[i][j] + s[i][k] * a.s[k][j] % mod) % mod;
            }
        }
        return ans;
    }
};

MAT pow_mat (int n) {
    MAT ans, x;
    ans.set();
    x.set();

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

    return ans;
}

int solve (int n) {
    if (n < 3)
        return n;
    else if (n == 3)
        return 6;
    else if (n == 4)
        return 19;

    MAT ans = pow_mat(n-5);
    return ((ans.s[2][0] * 2 + ans.s[2][1] * 6 + ans.s[2][2] * 19) % mod + mod) % mod;
}

int main () {
    int cas, n;
    scanf("%d", &cas);
    for (int i = 1; i <= cas; i++) {
        scanf("%d", &n);
        printf("Case %d: ", i);

        int ans = solve(n);
        if (n <= 9)
            printf("%d\n", ans);
        else
            printf("%04d\n", ans);
    }
    return 0;
}

uva 10743 - Blocks on Blocks(矩阵快速幂)