[POJ3734]Blocks（递推，矩阵快速幂）

题目链接：http://poj.org/problem?id=3734

题意：n个方块4个颜色，现在要求涂色，其中两种颜色的数量是偶数，问多少种情况。

考虑sta(i)为当前i块的时候的合法涂色种类数，则有：EE、OE(EO)、OO三种情况。

EEi = 2 * EE(i-1) + OE(EO)(i-1)

OOi = OE(EO)i + 2 * OO(i-1)

OE(EO)i = 2 * (OO(i-1) + OE(EO)(i-1) + EE(i-1))

抽取出系数，矩阵为

|2 1 0|

|0 1 2|

|2 2 2|

最终合法为EEn，则是第一行第一列。

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;

typedef long long LL;

const int mod = 10007;
const int maxn = 100;
typedef struct Matrix {
    int m[maxn][maxn];
    int r;
    int c;
    Matrix() {
        r = c = 0;
        memset(m, 0, sizeof(m));
    }
} Matrix;

Matrix mul(Matrix m1, Matrix m2, int mod) {
    Matrix ans = Matrix();
    ans.r = m1.r;
    ans.c = m2.c;
    for(int i = 1; i <= m1.r; i++) {
        for(int j = 1; j <= m2.r; j++) {
            for(int k = 1; k <= m2.c; k++) {
                if(m2.m[j][k] == 0) continue;
                ans.m[i][k] = ((ans.m[i][k] + m1.m[i][j] * m2.m[j][k] % mod) % mod) % mod;
            }
        }
    }
    return ans;
}

Matrix quickmul(Matrix m, int n, int mod) {
    Matrix ans = Matrix();
    for(int i = 1; i <= m.r; i++) ans.m[i][i]  = 1;
    ans.r = m.r;
    ans.c = m.c;
    while(n) {
        if(n & 1) ans = mul(m, ans, mod);
        m = mul(m, m, mod);
        n >>= 1;
    }
    return ans;
}

int n;

int main() {
//    freopen("in", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        Matrix A;
        A.r = A.c = 3;
        A.m[1][1] = 2; A.m[1][2] = 1; A.m[1][3] = 0;
        A.m[2][1] = 2; A.m[2][2] = 2; A.m[2][3] = 2;
        A.m[3][1] = 0; A.m[3][2] = 1; A.m[3][3] = 2;
        Matrix B = quickmul(A, n, mod);
        printf("%d
", B.m[1][1]);
    }
    return 0;
}