Happy Necklace (矩阵快速幂 + 递推 + 取模)

Posted 2021-01-10 ACM蒟蒻

题目链接

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

 

递推公式: f[n] = f[n - 1] + f[n - 3];

 

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 struct matrix{
 6     long long int mat[4][4];
 7 }A,B;
 8 matrix mul(matrix a,matrix b){
 9     matrix tmp;
10     memset(tmp.mat,0,sizeof(tmp.mat));
11     for(int i = 0;i < 4;i++)
12         for(int j = 0;j < 4;j++)
13             for(int k = 0;k < 4;k++){
14                 tmp.mat[i][j] += a.mat[i][k] * b.mat[k][j];
15                 tmp.mat[i][j] %= 1000000007;
16         }
17     return tmp;
18 }
19 matrix pow_mat(matrix a,long long int n){ //注意n 是long long int 类型,老写成int 类型
20     matrix ans;
21     int num = 0;
22     memset(ans.mat,0,sizeof(ans.mat));
23     for(int i = 0;i < 4;i++)
24         ans.mat[i][i] = 1;
25     while(n){
26         if(n & 1)
27             ans = mul(ans,a);
28         a = mul(a,a);
29         n >>= 1;
30     }
31     return ans;
32 }
33 int main(){
34     long long int t,n;
35     scanf("%lld",&t);
36     while(t--){
37         scanf("%lld",&n);
38         if(n == 2){
39             printf("3
");
40             continue;
41         }
42         else if(n == 3){
43             printf("4
");
44             continue;
45         }
46         else if(n == 4){
47             printf("6
");
48             continue;
49         }
50         else if(n == 5){
51             printf("9
");
52             continue;
53         }
54         else{
55             A.mat[0][0] = 3,A.mat[0][1] = 4,A.mat[0][2] = 6,A.mat[0][3] = 9;
56             B.mat[0][0] = 0,B.mat[0][1] = 0,B.mat[0][2] = 1,B.mat[0][3] = 0;
57             B.mat[1][0] = 1,B.mat[1][1] = 0,B.mat[1][2] = 0,B.mat[1][3] = 1;
58             B.mat[2][0] = 0,B.mat[2][1] = 1,B.mat[2][2] = 1,B.mat[2][3] = 0;
59             B.mat[3][0] = 0,B.mat[3][1] = 0,B.mat[3][2] = 0,B.mat[3][3] = 1;
60             B = pow_mat(B,n - 5);
61             printf("%lld
",mul(A,B).mat[0][3]);
62         }
63     }
64     return 0;
65 }

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