Blocks（矩阵快速幂）

最新推荐文章于 2022-07-12 23:14:50 发布

iteye_6881

最新推荐文章于 2022-07-12 23:14:50 发布

阅读量104

点赞数

CC 4.0 BY-SA版权

分类专栏： POJ

本文链接：https://blog.youkuaiyun.com/iteye_6881/article/details/82601508

POJ 专栏收录该内容

102 篇文章

订阅专栏

本文介绍如何使用矩阵快速幂解决染色问题，给定一定数量的格子，每格可选择四种颜色进行染色，求解在染色过程中红色和绿色块均为偶数的情况下的不同染色方案总数，并通过取模操作得到最终结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Blocks

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4036		Accepted: 1799

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

题意：

给出 T（1 ~ 100），代表有 T 个样例，后给出 N（1 ~ 10 ^ 9），代表有 N 个格子。给每个格子填色，可以填红蓝绿黄四种颜色，求染成红色和绿色同时为偶数的染色方案个数。结果模 10007。

思路：

矩阵快速幂。书上例题。设 ai 为同时为偶数的方法数，bi 为一个为偶数一个为奇数的方法数， ci 为同时为奇数的方法数，所以：

ai+1 = 2 x ai + bi ；

bi+1 = 2 x ai + 2 x bi + 2 x ci ；

ci+1 = bi + 2 x ci；

于是可以生成矩阵：

ai+1 2 1 0 ai ai 2 1 0 1

( bi+1 ) = ( 2 2 2 ) x （ bi ） -> ( bi ) = ( 2 2 2 ) ^ i X ( 0 )

ci+1 0 1 2 ci ci 0 1 2 0

最后得出来的矩阵中的 A [ 0 ] [ 0 ] 就是答案。

AC：

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

using namespace std;

typedef vector<int> vec;
typedef vector<vec> mat;

const int MOD = 10007;

mat mul (mat a, mat b) {
    mat c(a.size(), vec(b[0].size()));

    for (int i = 0; i < a.size(); ++i) {
        for (int j = 0; j < b[0].size(); ++j) {
            for (int k = 0; k < b.size(); ++k) {
                c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % MOD;
            }
        }
    }

    return c;
}

mat pow (mat a, int n) {
    mat b(a.size(), vec(a[0].size()));
    for (int i = 0; i < a.size(); ++i) {
        b[i][i] = 1;
    }

    while (n > 0) {
        if (n & 1) b = mul(b, a);
        a = mul(a, a);
        n >>= 1;
    }

    return b;
}

int main() {

    int t;
    scanf("%d", &t);

    while (t--) {
        int n;
        scanf("%d", &n);

        mat a(3, vec(3));
        a[0][0] = 2, a[0][1] = 1, a[0][2] = 0;
        a[1][0] = 2, a[1][1] = 2, a[1][2] = 2;
        a[2][0] = 0, a[2][1] = 1, a[2][2] = 2;

        a = pow(a, n);

        printf("%d\n", a[0][0]);
    }

    return 0;
}