hdu 3519 Lucky Coins Sequence-优快云博客

本文探讨了一种称为“幸运硬币序列”的问题，通过矩阵快速幂和二分法求解特定条件下硬币序列的数量。文章详细介绍了算法原理及其实现过程，并提供了完整的代码示例。

Lucky Coins Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 581 Accepted Submission(s): 301

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

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zhengfeng

矩阵快速幂 + 二分
题意：幸运硬币序列就是存在两个以上相邻的硬币正反面一样，那么就罗列仅有两个相邻的情况，最后用2^n减去那个值仅有两个相邻的状态相同的情况，如果从开始只有一个硬币开始，存在正反两种情况，然后就跟菲波那契数列一样了，现在我们就是要加硬币，可以加一枚硬币，也可以是加两枚（这个就是，正的时候加正反，反的时候加上反正，这样就恰好达到相邻只有两枚状态一样的结果了）所以这个就是菲波那契项。。。。最后的答案就是 2^n-Fibonacci(n+1) 这两部分用二分求，和快速幂求斐波那契项。 WA了好几次，原因是求2^n的二分里面，没有对2的幂取模，忘写了。。。wa的心疼。。

#include <cstdio>
#include <algorithm>
#include <iostream>
#define mod 10007
using namespace std;
struct Matrix
{
    int m[3][3];
};
Matrix E,B;
void Init()
{
    E.m[1][1]=E.m[2][2]=1;
    E.m[1][2]=E.m[2][1]=0;
    B.m[1][1]=B.m[1][2]=1;
    B.m[2][1]=1;
    B.m[2][2]=0;
}
Matrix Multi(Matrix A,Matrix B)
{
    Matrix ans;
    for(int i=1;i<=2;i++)
    for(int j=1;j<=2;j++)
    {
        ans.m[i][j]=0;
        for(int k=1;k<=2;k++)
        {
            ans.m[i][j]+=A.m[i][k]*B.m[k][j];
            ans.m[i][j]%=mod;
        }
    }
    return ans;
}
Matrix Pow(Matrix A,int k)
{
    Matrix ans=E;
    while(k)
    {
        if(k&1)
        {
            k--;
            ans=Multi(ans,A);
        }
        else
        {
            k/=2;
            A=Multi(A,A);
        }
    }
    return ans;
}
int POW(int a,int b)
{
    int ans=1;
    while(b)
    {
        if(b&1)
        {
            b--;
            ans*=a;
            ans%=mod;
        }
        else
        {
            b/=2;
            a*=a;
            a%=mod;
        }
    }
    return ans;
}
int p;
void solve()
{
    Matrix P=Pow(B,p);
    int ans=POW(2,p);
    ans-=(2*P.m[1][1])%mod;
    while(ans<0)
        ans+=mod;
    printf("%d\n",ans%mod);
}
int main()
{
    Init();
    while(scanf("%d",&p)!=EOF)
    {
        if(p==0)
            printf("0\n");
        else
            solve();
    }
    return 0;
}