hdoj2604Queuing【矩阵快速幂】

Queuing Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3694    Accepted Submission(s): 1666 Problem Description 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.   Input Input a length L (0 <= L <= 10 6) and M.   Output Output K mod M(1 <= M <= 30) where K is the number of E-queues with length L.   Sample Input 3 8 4 7 4 8   Sample Output 6 2 1   Author WhereIsHeroFrom   Source HDU 1st “Vegetable-Birds Cup” Programming Open Contest

递推公式：f
=f[n-1]+f[n-3]+f[n-4];

初始矩阵：1 1 0 0

                    0 0 1 0

                   1 0 0 1

                   1 0 0 0

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int a[5][5];
int ans[5][5];
int M;
void Mulit(int A[5][5],int B[5][5]){
int D[5][5]={0};
for(int i=0;i<4;++i){
for(int k=0;k<4;++k){
if(A[i][k]){
for(int j=0;j<4;++j){
D[i][j]=(D[i][j]+A[i][k]*B[k][j])%M;
}
}
}
}
for(int i=0;i<4;++i){
for(int j=0;j<4;++j){
A[i][j]=D[i][j];
}
}
}
void Matrix(int A[5][5],int n){
memset(ans,0,sizeof(ans));
ans[0][0]=ans[1][1]=ans[2][2]=ans[3][3]=1;
while(n){
if(n&1)Mulit(ans,a);
Mulit(a,a);
n>>=1;
}
}
int main()
{
int n,i,j,k;
int num[5]={1,2,4,6,9};
while(scanf("%d%d",&n,&M)!=EOF){
if(n<=4){
printf("%d\n",num
%M);
}
else {
memset(a,0,sizeof(a));
a[0][0]=1;a[2][0]=1;a[3][0]=1;
a[0][1]=1;a[1][2]=1;a[2][3]=1;
Matrix(a,n-4);
int S=9*ans[0][0]+6*ans[1][0]+4*ans[2][0]+2*ans[3][0];
printf("%d\n",S%M);
}
}
return 0;
}