HDU 6030 Happy Necklace【矩阵快速幂】

Happy Necklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 477    Accepted Submission(s): 198

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中国大学生程序设计竞赛 - 女生专场

由于2是最小的素数，考虑长度为2的子串。红色为A，蓝色为B，则只有AA，AB，BA三种状况。对每种状况，在后面加上A或B，AA能够造成AA，AB，AB能够造成BA，BA能够造成AA。经过这个递推扩展到长度为n的状况，用矩阵快速幂加速便可。矩阵为：

初始状况下，AA，AB，BA都有可能，所以最后将矩阵中的全部数字相加就是答案。

注意n为10的18次方会爆int因此在矩阵快速幂的时候要用long long【Matrix quickpow(Matrix A,ll k)】

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>

#define INF 0x3f3f3f3f

#define mod 1000000007

using namespace std;

typedef long long ll;
const int maxn = 100010;

ll n;

struct Matrix {
	ll a[5][5];
};


Matrix mul(Matrix x, Matrix y)
{
	Matrix temp;
	for (int i = 1; i <= 3; i++)
		for (int j = 1; j <= 3; j++) temp.a[i][j] = 0;

	for (int i = 1; i <= 3; i++)
	{
		for (int j = 1; j <= 3; j++)
		{
			ll sum = 0;
			for (int k = 1; k <= 3; k++)
			{
				sum = (sum + x.a[i][k] * y.a[k][j] % mod) % mod;
			}
			temp.a[i][j] = sum;
		}
	}
	return temp;
}

Matrix quickpow(Matrix A,ll k)
{
	Matrix res;
	res.a[1][1] = 1; res.a[1][2] = 0; res.a[1][3] = 0;
	res.a[2][1] = 0; res.a[2][2] = 1; res.a[2][3] = 0;
	res.a[3][1] = 0; res.a[3][2] = 0; res.a[3][3] = 1;
	while (k)
	{
		if (k & 1) res = mul(res, A);
		A = mul(A, A);
		k >>= 1;
	}
	return res;
}

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%lld", &n);
		if (n == 2)
		{
			printf("3\n");
			continue;
		}
		Matrix A;
		A.a[1][1] = 1; A.a[1][2] = 0; A.a[1][3] = 1;
		A.a[2][1] = 1; A.a[2][2] = 0; A.a[2][3] = 0;
		A.a[3][1] = 0; A.a[3][2] = 1; A.a[3][3] = 0;
		Matrix res = quickpow(A, n - 2);
		ll x = (res.a[1][1] + res.a[1][2] + res.a[1][3]) % mod;
		ll y = (res.a[2][1] + res.a[2][2] + res.a[2][3]) % mod;
		ll z = (res.a[3][1] + res.a[3][2] + res.a[3][3]) % mod;
		printf("%lld\n", (x + y + z) % mod);
	}
}

1018
101Matrix quickpow(Matrix A,ll k)】1018101810181018101810181018101810

10 18

10 18

10 18

10 18