HDU6029 Happy Necklace 2017-05-07 19:11 45人閱讀 評論(0) 收藏

Happy Necklace

                                                                          Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
                                                                                                          Total Submission(s): 19    Accepted Submission(s): 9

Problem Description

Little Q wants to buy a necklace for his girlfriend. Necklaces are single strings composed of multiple red and blue beads.
Little Q desperately wants to impress his girlfriend, he knows that she will like the necklace only if for every prime length continuous subsequence in the necklace, the number of red beads is not less than the number of blue beads.
Now Little Q wants to buy a necklace with exactly  n  beads. He wants to know the number of different necklaces that can make his girlfriend happy. Please write a program to help Little Q. Since the answer may be very large, please print the answer modulo  109+7 .
Note: The necklace is a single string,  {not a circle}.

Input

The first line of the input contains an integer  T(1≤T≤10000) , denoting the number of test cases.
For each test case, there is a single line containing an integer  n(2≤n≤1018) , denoting the number of beads on the necklace.

Output

For each test case, print a single line containing a single integer, denoting the answer modulo  109+7 .

Sample Input

  
  
  
  

   
   
   
   2
2
3
  
  
  
  

 

Sample Output

  
  
  
  

   
   
   
   3
4
  
  
  
  

 

Source

2017中國大學生程序設計競賽 - 女生專場

 

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題意：一個項鍊有n個珠子，這個項鍊是個鏈，不是一個環，項鍊上連續素數個珠子中紅色珠子個數要大於等於藍色珠子

解題思路：找出前面幾個，發現存在遞推關係f[i]=f[i-1]+f[i-3],而後構造矩陣，矩陣快速冪

                                                                                   {1 1 0}

     矩陣構造: {a[i],a[i-1],a[i-2]}={a[i-1],a[i-2],a[i-3]}*{0 0 1}

                                                                                   {1 0 0}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;
#define mod 1000000007

LL n;

struct Matrix
{
    LL v[5][5];
    Matrix()
    {
        memset(v,0,sizeof v);
    }
}dan;

Matrix mul(Matrix a,Matrix b,int d)
{
    Matrix ans;
    for(int i=1; i<=d; i++)
    {
        for(int j=1; j<=d; j++)
        {
            for(int k=1; k<=d; k++)
            {
                ans.v[i][j]+=a.v[i][k]*b.v[k][j];
                ans.v[i][j]%=mod;
            }
        }
    }
    return ans;
}

Matrix pow(Matrix a,LL k,int d)
{
    Matrix ans=dan;
    while(k)
    {
        if(k&1) ans=mul(ans,a,d);
        k>>=1;
        a=mul(a,a,d);
    }
    return ans;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld",&n);
        if(n==2) {printf("3\n");continue;}
        if(n==3) {printf("4\n");continue;}
        if(n==4) {printf("6\n");continue;}
        Matrix ans,a;
        a.v[1][1]=a.v[3][1]=a.v[1][2]=a.v[2][3]=1;
        dan.v[1][1]=6,dan.v[1][2]=4,dan.v[1][3]=3;
        ans=pow(a,n-4,3);
        printf("%lld\n",ans.v[1][1]);
    }
    return 0;
}