HDU 2604 Queuing

Queuing

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 690    Accepted Submission(s): 337

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 2^L 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

//本題的思想是推出公式，F（N）=F（N-1）+F（N-3）+F（N-4） //因爲假設F（N）爲已經是E隊列的個數， 那麼有3種情況 //就是一，在F（N-1）後加一個M //二，在F（N-3）後加MMF //三，在F（N-4）後加MMFF //所以得到遞推式後即可用 矩陣乘法 解決問題 #include<stdio.h> struct node { long long num[4][4]; }; node res,tp; int n,m; node mut(node a,node b) { node c; int i,j,k; for(i=0;i<4;i++) for(j=0;j<4;j++) { c.num[i][j]=0; for(k=0;k<4;k++) c.num[i][j]+=(a.num[i][k]*b.num[k][j])%m; c.num[i][j]%=m; } return c; } node abc(int temp,node p) { tp=p; p=res; while(temp) { if(temp&1) p=mut(p,tp); tp=mut(tp,tp); temp>>=1; } return p; } int main() { int i,j; for(i=0;i<4;i++) for(j=0;j<4;j++) res.num[i][j]=(i==j); node p; p.num[0][0]=p.num[0][2]=p.num[0][3]=p.num[1][0]=p.num[2][1]=1; p.num[3][2]=1; p.num[1][1]=p.num[1][2]=p.num[1][3]=p.num[2][0]=0; p.num[0][1]=p.num[2][2]=p.num[2][3]=p.num[3][0]=0; p.num[3][1]=p.num[3][3]=0; int a[5]={0,2,4,6,9}; while(scanf("%d%d",&n,&m)!=EOF) { if(n<=4) { printf("%d/n",a[n]%m); continue; } node k; k=abc(n-4,p); int sum=(k.num[0][0]*9+k.num[0][1]*6+k.num[0][2]*4+k.num[0][3]*2)%m; printf("%d/n",sum); } return 0; }

下面的一樣

#include<stdio.h> int n,m,l; long long a[4][4],sum[4][4],hash[4][4]; void find() { int i,j,p,t,k; for(p=n-4;p>0;) { if(p&1) { for(i=0;i<4;i++) { for(j=0;j<4;j++) { for(k=0,t=0;k<4;k++) t=t+sum[i][k]*hash[k][j]; a[i][j]=t%m; } } for(i=0;i<4;i++) for(j=0;j<4;j++) hash[i][j]=a[i][j]; p--; } else { for(i=0;i<4;i++) { for(j=0;j<4;j++) { for(k=0,t=0;k<4;k++) t=t+sum[i][k]*sum[k][j]; a[i][j]=t%m; } } for(i=0;i<4;i++) for(j=0;j<4;j++) sum[i][j]=a[i][j]; p=p>>1; } } } int main() { int i,j; int b[5]={0,2,4,6,9}; while(scanf("%d%d",&n,&m)!=EOF) { if(n<5) { printf("%d/n",b[n]%m); continue; } for(i=0;i<4;i++) for(j=0;j<4;j++) sum[i][j]=hash[i][j]=0; for(i=0;i<4;i++) hash[i][i]=1; sum[0][0]=sum[0][2]=sum[0][3]=1; sum[1][0]=sum[2][1]=sum[3][2]=1; find(); l=a[0][0]*9+a[0][1]*6+a[0][2]*4+a[0][3]*2; l=l%m; printf("%d/n",l); } return 0; }

還可以用打表法

/************************************************************************ 前四位我們可以算出d[1]=2,d[2]=4,d[3]=6,d[4]=9. 我們可以這樣想：一個合法串可以由兩個較短的合法串組成 就以d[n]爲例：(注意不能重複) 1、n-1個字符的時候: +m 2、n-2: 只能+mm，會和n-1重複,所以不考慮n-2 3、n-3: +mmf 4、n-4: +mmff 5、n-5: 如果是+mmffm，會與n-1重複,+mmmff會與n-4重複,+mmmmf會與n-3重複(不考慮) 所以就考慮n-1,n-3,n-4，DP等式就出來了:dp[n]=dp[n-1]+dp[n-3]+dp[n-4] n<10^6 m<=30 用打表可以做 [剛剛開始直接用int敲的代碼，果斷MLE了。。。改成char試試 - -就過了] ************************************************************************/ #include <stdio.h> #define N 1000001 #define M 30 char node[N][M]; void init() { int i,j; for(i=1;i<=M;i++) { node[1][i]=2%i; node[2][i]=4%i; node[3][i]=6%i; node[4][i]=9%i; for (j=5;j<=N;j++) { node[j][i]=(node[j-1][i]+node[j-3][i]+node[j-4][i])%i; } } } int main() { init(); int n,m; while (~scanf("%d%d",&n,&m)) { printf("%d/n",node[n][m]); } return 0; }