POJ 3734 Blocks(矩阵快速幂加递推)

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Blocks
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 6133   Accepted: 2931

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
  
  题目大意就是给你n个格子, n <= 10^9 , 每个格子可以涂红,蓝, 绿, 黄四种颜色, 但是必须要保证红色和绿色的数量为偶数。 问有几种涂的方案, 答案膜10007

  我的第一想法是递推, 但是10^9实在是太大了, 打出一个表大约在6秒左右, 所有这时候就应该想到矩阵快速幂。
  首先看状态, 当n == 1时, 很明显只能有两种方案。 这样我们不妨用二维数组dp[n][i] 来表示当前的方案数, N 为格子的数量, 0 <= i <= 3

  
   1. dp[N][0] :表示N块中红色绿色的数量均为偶数。
   2. dp[N][1] :表示N块中红色为偶数,绿色为奇数。
   3. dp[N][2] :表示N块中红色为奇数,绿色为偶数。
   4. dp[N][3] :表示N块中红色绿色的数量均为奇数。
  
  则dp转移方程就为:

   dp[N+1][0] = 2 * dp[N][0] + 1 * dp[N][1] + 1 * dp[N][2] + 0 * dp[N][3]

   dp[N+1][1] = 1 * dp[N][0] + 2 * dp[N][1] + 0 * dp[N][2] + 1 * dp[N][3]

   dp[N+1][2] = 1 * dp[N][0] + 0 * dp[N][1] + 2 * dp[N][2] + 1 * dp[N][3]

   dp[N+1][3] = 0 * dp[N][0] + 1 * dp[N][1] + 1 * dp[N][2] + 2 * dp[N][3]
  仔细分析就不难看明白转移方程。
   上述的转移方程能够转化为矩阵:

   |2 1 1 0|

   |1 2 0 1|

   |1 0 2 1|

   |0 1 1 2|
   
   还有一个坑点就是实现快速幂的时候, 应该将结果矩阵初始化为单位矩阵, 否则会让一个0矩阵一直乘啊乘, 结果当然是0.
   最后再说一下我对矩阵的理解, 很明显, dp[N][i] 只有当i == 0 的时候时是符合题意的, 当矩阵的幂为1的时候,这时候只有一块砖, 所以只有两种方案, 当矩阵的幂为2的时候, 首先要看矩阵幂为1时候的状态, 再由i 的四种状态决定最后一块砖到底涂什么颜色, 不管怎么说, 答案总是当i == 0 时且 n == 0即dp[0][0] 总为答案。
 
   
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

typedef long long LL;
const int MOD = 10007;

struct Matrix {
    LL a[4][4];
    int n, m;
    void clear() {
        n = m = 0;
        memset( a, 0, sizeof(a) );
    }
    Matrix operator * ( const Matrix b ) {
        Matrix res;
        res.clear();
        res.n = n;
        res.m = b.m;
        for( int i = 0; i < n; i++ ) {
            for( int j = 0; j < b.m; j++ ) {
                for( int k = 0; k < m; k++ ) {
                    res.a[i][j] += a[i][k] * b.a[k][j];
//                    printf( "==========\\n" );
                }
                res.a[i][j] %= MOD;
            }
        }
        return res;
    }
};

Matrix quick_power( Matrix a, int n ) {
    
    Matrix res;
    memset( res.a, 0, sizeof(res.a) );
    res.clear();
    res.n = res.m = 4;
    for( int i = 0; i < 4; i++ ) {
        res.a[i][i] = 1;
    }
    while( n ) {
        if( n & 1 ) res = res * a;
        
        n >>= 1;
        a = a * a;
       
    }
    return res;
}

int main() {
    int t;
    scanf( "%d", &t );
    Matrix A;
    A.clear();
    A.n = A.m = 4;
    A.a[0][0] = 2; A.a[0][1] = 1; A.a[0][2] = 1; A.a[0][3] = 0;
    A.a[1][0] = 1; A.a[1][1] = 2; A.a[1][2] = 0; A.a[1][3] = 1;
    A.a[2][0] = 1; A.a[2][1] = 0; A.a[2][2] = 2; A.a[2][3] = 1;
    A.a[3][0] = 0; A.a[3][1] = 1; A.a[3][2] = 1; A.a[3][3] = 2;
    while( t-- ) {
        int n;
        scanf( "%d", &n );
        Matrix res = quick_power( A, n );
        printf( "%lld\\n", res.a[0][0] );
    }
    return 0;
}
View Code

 

    这道题很有意义, 果然学计算机还是要学好数学!
 
   加油, stick on it!
  

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