ACM_HDUOJ_2604_Queuing排队问题

Queuing

[b]Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4554    Accepted Submission(s): 2002
[/b]

[align=left]Problem Description[/align]
Queues and Priority Queues are data structures which are known to most computer scientists. The Queue occurs often in our daily life. There are many people lined up at the lunch time.



  Now we define that ‘f’ is short for female and ‘m’ is short for male. If the queue’s length is L, then there are 2L numbers of queues. For example, if L = 2, then they are ff, mm, fm, mf . If there exists a subqueue
as fmf or fff, we call it O-queue else it is a E-queue.

Your task is to calculate the number of E-queues mod M with length L by writing a program.

 

[align=left]Input[/align]
Input a length L (0 <= L <= 10
6) and M.
 

[align=left]Output[/align]
Output K mod M(1 <= M <= 30) where K is the number of E-queues with length L.
 

[align=left]Sample Input[/align]

3 8
4 7
4 8

 

[align=left]Sample Output[/align]

6
2
1问题描述 两个参数L（队列的长度）和M（结果对M取余）。计算队列中不含有fff且不含有fmf的队列个数对M取余。问题分析 l=0->0;l=1->2;l=2->4;l=3->6;l=4->9;l=5->15; 找规律可知F(l)=F(l-1)+F(l-3)+F(l-4);可构造矩阵 |1，0，1，1|与矩阵|F(n-1)|相乘得|F(n) |由此可知 |1，0，0，0| |F(n-2)| |F(n-1)| |0，1，0，0| |F(n-3)| |F(n-2)| |0，0，1，0| |F(n-4)| |F(n-3)|F(n)是矩阵|1,0,1,1|的(n-4)次幂与矩阵|F(4)|相乘的第一行第一列元素 |1,0,0,0| |F(3)| |0,1,0,0| |F(2)| |0,0,1,0| |F(1)|用快速幂对矩阵求n次幂即可；AC代码

#include <cmath>
#include<iostream>
//#include<algorithm>
#include<cstring>
using namespace std;
int m;
struct Matrix{
long long int m_matrix[4][4];
Matrix operator*(const Matrix& rhs)const{
Matrix temp;
for(int i = 0 ; i < 4 ; i++){
for(int j = 0 ; j < 4 ; j++){
temp.m_matrix[i][j] = 0;
for(int k = 0 ; k < 4 ; k++){
temp.m_matrix[i][j] += (m_matrix[i][k])*(rhs.m_matrix[k][j])%m;
temp.m_matrix[i][j] %= m;
}
}
}
return temp;
}
};

int get(int l){
int n =l- 4;
Matrix rhs={
1,0,1,1,
1,0,0,0,
0,1,0,0,
0,0,1,0};
Matrix ans;
memset(ans.m_matrix , 0 , sizeof(ans.m_matrix));
for(int i = 0 ; i < 4 ; i++)
ans.m_matrix[i][i] = 1;
while(n){
if(n&1)
ans = ans*rhs;
n >>= 1;
rhs = rhs*rhs;
}
int sum = 0;
sum += ans.m_matrix[0][0]*9%m;
sum += ans.m_matrix[0][1]*6%m;
sum += ans.m_matrix[0][2]*4%m;
sum += ans.m_matrix[0][3]*2%m;
return sum%m;
}

int main()
{
int l,s;
while(cin>>l>>m){
switch(l){
case 0:s=0;break;
case 1:s=2;break;
case 2:s=4;break;
case 3:s=6;break;
case 4:s=9;break;
default:break;
}

if(l>=5){

s=get(l);
}
cout<<s%m<<endl;
}
return 0;
}