poj3519 Lucky Coins Sequence矩阵快速幂

Lucky Coins Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 608 Accepted Submission(s): 319

Problem Description

As we all know,every coin has two sides,with one side facing up and another side facing down.Now,We consider two coins's state is same if they both facing up or down.If we have N coins and put them in a line,all of us know that it will be 2^N different ways.We call a "N coins sequence" as a Lucky Coins Sequence only if there exists more than two continuous coins's state are same.How many different Lucky Coins Sequences exist?

 

Input

There will be sevaral test cases.For each test case,the first line is only a positive integer n,which means n coins put in a line.Also,n not exceed 10^9.

 

Output

You should output the ways of lucky coins sequences exist with n coins ,but the answer will be very large,so you just output the answer module 10007.

 

Sample Input

3 4

 

Sample Output

2 6

 

Source

2010 ACM-ICPC Multi-University Training Contest（9）——Host by HNU 

分析题意，我们可以发现就是dp,怎么求递推公式呢？dp[i][j]表示，有i位长，j最后几位是相连的！

dp[i][3]=dp[i-1][2];

dp[i][2]=dp[i-1][1];

dp[i][1]=dp[i-1][1]+dp[i-1][2];

dp[1][1]=2;dp[1][2]=0;dp[1][3]=0;

这样，我们就可以转化为矩阵求和了！

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define mod 10007

struct node {
   int  m[4][4];
    node operator *(node b) const//重载乘法
    {
        int i,j,k;
        node c;
        for(i=0;i<4;i++)
            for(j=0;j<4;j++)
            {
                c.m[i][j]=0;
                for(k=0;k<4;k++)
                {
                    c.m[i][j]+=m[i][k]*b.m[k][j];
                    c.m[i][j]%=mod;//都要取模
                }
            }
        return c;
    }
};
node original,result;
void quickm(int n)
{
    node a,b;
    b=original;a=result;
    while(n)
    {
        if(n&1)
        {
          b=b*a;
        }
        n=n>>1;
        a=a*a;
    }
    printf("%d
",2*b.m[0][3]%mod);
}
int main ()
{

    int i,j,n;
    for(i=0;i<4;i++)
        for(j=0;j<4;j++)
        {
            original.m[i][j]=(i==j)?1:0;//初始化为单位矩阵
        }
    memset(result.m,0,sizeof(result.m));
    result.m[0][0]=result.m[0][1]=result.m[1][0]=result.m[1][2]=result.m[2][3]=1;
    result.m[3][3]=2;
    while(scanf("%d",&n)!=EOF)
    {

        quickm(n);
    }
    return 0;
}