HDU 6030 Happy Necklace【矩陣快速冪】

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中國大學生程序設計競賽 - 女生專場

由於2是最小的素數，考慮長度爲2的子串。紅色爲A，藍色爲B，則只有AA，AB，BA三種狀況。對每種狀況，在後面加上A或B，AA能夠造成AA，AB，AB能夠造成BA，BA能夠造成AA。經過這個遞推擴展到長度爲n的狀況，用矩陣快速冪加速便可。矩陣爲：

初始狀況下，AA，AB，BA都有可能，所以最後將矩陣中的全部數字相加就是答案。

注意n爲10的18次方會爆int因此在矩陣快速冪的時候要用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