poj 3734 Blocks【矩阵快速幂染色】

最新推荐文章于 2023-12-18 00:59:23 发布

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本文探讨了一个有趣的数学问题：如何计算在特定条件下（红色和绿色方块的数量必须为偶数），一条由N个方块组成的线可以有多少种不同的涂色方式（每种方块可以选择红、蓝、绿、黄四种颜色之一）。文章提供了一种使用矩阵快速幂的方法来高效解决这个问题。

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Description

Panda has received an assignment of painting a line of blocks. Since Panda is such an intelligent boy, he starts to think of a math problem of painting. Suppose there are N blocks in a line and each block can be paint red, blue, green or yellow. For some myterious reasons, Panda want both the number of red blocks and green blocks to be even numbers. Under such conditions, Panda wants to know the number of different ways to paint these blocks.

Input

The first line of the input contains an integer T(1≤T≤100), the number of test cases. Each of the next T lines contains an integer N(1≤N≤10^9) indicating the number of blocks.

Output

For each test cases, output the number of ways to paint the blocks in a single line. Since the answer may be quite large, you have to module it by 10007.

Sample Input

2
1
2
Sample Output

2
6
思路：
a[i]：表示前i个红绿都是偶数的方案
b[i]：表示前i个红绿其中有一个是偶数的方案
c[i]：表示前i个红绿都不是偶数的方案
a[i + 1] = 2 * a[i] + b[i] ;
b[i + 1] = 2 * a[i] + 2 * b[i] + 2 * c[i] ;
c[i + 1] = b[i] + 2 * c[i] ;

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int mod = 10007;

struct mat {
    int mapp[3][3];
}base, ans;

mat mat_mul(mat A, mat B) {
    mat C;
    memset(C.mapp, 0, sizeof(C.mapp));
    for(int i = 0; i < 3; i++) {
        for(int j = 0; j < 3; j++) {
            for(int k = 0; k < 3; k++) {
                C.mapp[i][k] = (C.mapp[i][k] + A.mapp[i][j] * B.mapp[j][k] % mod) % mod;
            }
        }
    }
    return C;
}

int quick_pow(mat A, int b) {
    memset(ans.mapp, 0, sizeof(ans.mapp));
    for(int i = 0; i < 3; i++) {
        ans.mapp[i][i] = 1; //初始矩阵
    }
    while(b) {
        if(b & 1) ans = mat_mul(ans, A);
        A = mat_mul(A, A);
        b >>= 1;
    }
    return ans.mapp[0][0];
}

int main() {
    int t, n;
    scanf("%d", &t);
    while(t--) {
        base.mapp[0][0] = 2; //推导矩阵
        base.mapp[0][1] = 1;
        base.mapp[0][2] = 0;
        base.mapp[1][1] = 2;
        base.mapp[1][2] = 2;
        base.mapp[2][1] = 1;
        base.mapp[2][2] = 2;
        scanf("%d", &n);
        printf("%d\n", quick_pow(base, n) % mod);
    }
    return 0;
}