Queuing
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4638 Accepted Submission(s): 2045
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.
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
Recommend
lcy
对于长度为n的队列q(n),做以下讨论。
图中 ,".....m"表示以字符'm'结尾的queue,其它类似。E_q表示E_queue。
定义f(n)为队列q(n)之中满足E_queue的队列个数。
可得f(n)=f(n-1)+f(n-3)+f(n-4)
转换为矩阵,运用矩阵快速幂运算进行求解。
即,
n>=4,
1 #include <iostream> 2 #include <algorithm> 3 #include <map> 4 #include <vector> 5 #include <functional> 6 #include <string> 7 #include <cstring> 8 #include <queue> 9 #include <set> 10 #include <cmath> 11 #include <cstdio> 12 using namespace std; 13 #define IOS ios_base::sync_with_stdio(false)] 14 typedef long long LL; 15 const int INF=0x3f3f3f3f; 16 17 const int maxn=6; 18 int modnum; 19 typedef struct matrix{ 20 int v[maxn][maxn]; 21 void init(){memset(v,0,sizeof(v));} 22 }M; 23 M mul_mod(const M &a,const M &b,int L,int m,int n) 24 { 25 M c; c.init(); 26 for(int i=0;i<L;i++){ 27 for(int j=0;j<n;j++){ 28 for(int k=0;k<m;k++)//注意j,k范围 29 c.v[i][j]+=a.v[i][k]*b.v[k][j]%modnum; 30 c.v[i][j]%=modnum; 31 } 32 } 33 return c; 34 } 35 M power(M x,int L,int p) 36 { 37 M tmp; tmp.init(); 38 for(int i=0;i<L;i++) 39 tmp.v[i][i]=1; 40 while(p){ 41 if(p&1) tmp=mul_mod(x,tmp,L,L,L); 42 x=mul_mod(x,x,L,L,L); 43 p>>=1; 44 } 45 return tmp; 46 } 47 int main() 48 { 49 int L; 50 M a,b; 51 while(scanf("%d%d",&L,&modnum)!=EOF){ 52 b.init(); 53 b.v[0][0]=9%modnum; 54 b.v[1][0]=6; 55 b.v[2][0]=4; 56 b.v[3][0]=2; 57 if(L==0){printf("1 "); continue;} 58 if(L<5){printf("%d ",b.v[4-L][0]); continue;} 59 a.init(); 60 a.v[0][0]=a.v[0][2]=a.v[0][3]=a.v[1][0]=a.v[2][1]=a.v[3][2]=1; 61 a=power(a,4,L-4); 62 a=mul_mod(a,b,4,4,1); 63 printf("%d ",a.v[0][0]); 64 } 65 }
网上有介绍用trie图得到递推式的方法,根节点为字符串的最后一个字符,根据前一个字符进行状态转移。
不是非常理解,暂且记录下来。