【解題報告】hdu6030 Happy Necklace

Happy Necklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 863    Accepted Submission(s): 383

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

　

【題目大意】用紅藍兩種珠子，串成一條項鍊，

使得任意素數長度的區間都有紅色珠子大於等於藍色珠子，問有多少種串法

 

 

【思路】考慮最小的素數2，任意素數長度區間 L = 2k + 1, 所以只要任意長度爲2和3的區間知足條件便可

假設 1 = 紅， 0 = 藍　　

易得 F( 2 ) = 3,

F（3） = 4

要求長度爲n的串，能夠在n-1的串後加1 或 0,

加 1 確定知足要求

加 0 要知足 n-1 串的末尾不能是0,且倒數第二位也不能是0，所以在n-3串後直接加110

得遞推式 F（n） = F（n-1） + F（n-3） 

-------------------------------------------------------------------------

再用 矩陣快速冪 加速求遞推式

 

 求得 A =

       

 

【AC代碼】

 

/*
矩陣快速冪模板 
by chsobin
*/
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn = 3;
const ll mod = 1e9 + 7;

//矩陣結構體 
struct Matrix{
    ll a[maxn][maxn];
    void init(){    //初始化爲單位矩陣 
        memset(a, 0, sizeof(a));
        for(int i=0;i<maxn;++i){
            a[i][i] = 1;
        }
    }
};

//矩陣乘法 
Matrix mul(Matrix a, Matrix b){
    Matrix ans;
    for(int i=0;i<maxn;++i){
        for(int j=0;j<maxn;++j){
            ans.a[i][j] = 0;
            for(int k=0;k<maxn;++k){
                ans.a[i][j] += a.a[i][k] * b.a[k][j];
                ans.a[i][j] %= mod; 
            }
        }
    } 
    return ans;
}

//矩陣快速冪 
Matrix qpow(Matrix a, ll n){
    Matrix ans;
    ans.init();
    while(n){
        if(n&1) ans = mul(ans, a);
        a = mul(a, a);
        n /= 2;
    } 
    return ans;
}

////for debug
//void output(Matrix a){
//    for(int i=0;i<maxn;++i){
//        for(int j=0;j<maxn;++j){
//            cout << a.a[i][j] << " ";
//        }
//        cout << endl;
//    }
//}


int main(){
    Matrix A = { 1,1,0,0,0,1,1,0,0 };    //遞推矩陣 
    
    int t;
    cin >> t;
    ll n;
    int ans[5] = {0, 0, 3, 4, 6}; //枚舉初始狀況 
    while(t--){
        cin >> n;
        if(n<=4) cout << ans[n] << endl;
        else{
            Matrix m = qpow(A , n-4);
            ll temp = 6 * m.a[0][0] % mod;
            temp = (temp + 4 * m.a[1][0]%mod)%mod;
            temp = (temp + 3 * m.a[2][0]%mod)%mod;
            cout << temp << endl;
        }
    }
    return 0;
}

AC

 

【總結】：

         看到數據量就明白了這一類題的作法

         1.打表

         2.根據打表結果，寫出遞推方程。

         3.將遞推方程構造爲矩陣。

         4.快速冪求解。