poj 3734 Blocks (递推，矩阵快速幂)

原来矩阵还能解决这样的题。。真是长知识了

让我们试着从左边开始染色。设染到第i个方块，红绿都是偶数的方案数是ai，红绿方块数为一奇一偶的方案数是bi，红绿方块数都为奇数的方案数是ci；

我们就可以得到这样的递推式：

a(i+1) = 2*ai + bi

b(i+1) = 2*ai + 2*bi + 2*ci

c(i+1) = bi + 2*ci

（a0 = 1, b0 = 0, c0 = 0）

所以就可以构造出矩阵了。。

代码:

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;
const int mod = 1e4+7;
struct node
{
    ll s[3][3];
};

node mul(node a, node b)
{
    node t;
    memset(t.s, 0, sizeof(t.s));
    for(int i = 0; i < 3; i++)
        for(int j = 0; j < 3; j++)
            for(int k = 0; k < 3; k++)
                t.s[i][j] = (t.s[i][j]+a.s[i][k]*b.s[k][j])%mod;
    return t;
}

node mt_pow(node p, int k)
{
    node q;
    memset(q.s, 0, sizeof(q.s));
    for(int i = 0; i < 3; i++)
        q.s[i][i] = 1;
    while(k)
    {
        if(k&1) q = mul(p, q);
        p = mul(p, p);
        k /= 2;
    }
    return q;
}

int main(void)
{
    int n, t;
    cin >> t;
    while(t--)
    {
        scanf("%d", &n);
        if(!n) puts("1");
        else
        {
            node base;
            base.s[0][0] = 2; base.s[0][1]  = 1; base.s[0][2] = 0;
            base.s[1][0] = 2; base.s[1][1]  = 2; base.s[1][2] = 2;
            base.s[2][0] = 0; base.s[2][1]  = 1; base.s[2][2] = 2;
            node ans = mt_pow(base, n);
            printf("%d\n", ans.s[0][0]);
        }
    }
    return 0;
}

Blocks

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6319		Accepted: 3035

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

Source

PKU Campus 2009 (POJ Monthly Contest – 2009.05.17), Simon