POJ 3734-Blocks

最新推荐文章于 2019-10-05 22:48:21 发布

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本文深入探讨了快速幂算法在解决组合数学问题中的应用，特别关注了一个关于涂色块的问题。通过找到递推方程并利用矩阵乘法进行快速求解，展示了巧妙的算法设计与数学技巧。代码实现提供了具体的示例，帮助读者理解和实践这一解决方案。

Blocks

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4494		Accepted: 2038

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

小结：

今天最后一篇博客也该落幕了，快速幂算法也感觉有点理解了。

这类题目的关键是找到递推方程，然后通过线性代数矩阵的乘法来快速求出结果，方法十分巧妙，最初想到这种算法的人真的是个天才！

以下是AC代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;
typedef vector<int> vec;
typedef vector<vec> mat;
typedef long long ll;
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 k=0;k<b.size();k++)
    for(int j=0;j<b[0].size();j++)
    {
        c[i][j]=(c[i][j]+a[i][k]*b[k][j])%mod;
    }
    return c;
}
mat pow_mul(mat a,ll num)
{
    mat b(a.size(),vec(a[0].size()));
    for(int i=0;i<a.size();i++)
    b[i][i]=1;
    while(num)
    {
        if(num&1)
        b=mul(b,a);
        a=mul(a,a);
        num>>=1;
    }
    return b;
}
int main()
{
    ll n;
    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        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;
        scanf("%lld",&n);
        a=pow_mul(a,n);
        printf("%d\n",a[0][0]);
    }
    return 0;
}