HDU 2604（矩阵 / DP）

Queuing

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 540    Accepted Submission(s): 273

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 84 74 8

 

 

Sample Output

621

 DP打表做法：

/************************************************************************ 前四位我们可以算出d[1]=2,d[2]=4,d[3]=6,d[4]=9.我们可以这样想：一个合法串可以由两个较短的合法串组成就以d[n]为例：(注意不能重复)1、n-1个字符的时候: +m2、n-2: 只能+mm，会和n-1重复,所以不考虑n-23、n-3: +mmf4、n-4: +mmff5、n-5: 如果是+mmffm，会与n-1重复,+mmmff会与n-4重复,+mmmmf会与n-3重复(不考虑)所以就考虑n-1,n-3,n-4，DP等式就出来了:dp[n]=dp[n-1]+dp[n-3]+dp[n-4]n<10^6 m<=30 用打表可以做 [刚刚开始直接用int敲的代码，果断MLE了。。。改成char试试 - -|| 就过了]************************************************************************/#include <iostream>using namespace std;#define N 1000005#define M 30char 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;}

 

 

矩阵做法：

/************************************************************************ 这题的矩阵写法我悲剧了 - -|| MOD在全局定义了，不小心在main里面又定义了，RE了10多次说除0了。。。找到后一直WA~~~ 这个更无语，我数组才定义了4个长度。。。。。我太阳！！！！看来我是傻到家了！！！！！！！************************************************************************/#include <iostream>using namespace std;#define N 4struct Mat { int martix[N][N];};Mat res,q,tp,tp1;int MOD;int elem[5];Mat Martix_Mul(Mat a,Mat b){ int i,j,k; Mat c; for (i=0;i<N;i++) { for (j=0;j<N;j++) { c.martix[i][j]=0; for (k=0;k<N;k++) {c.martix[i][j]=(c.martix[i][j]+(a.martix[i][k]*b.martix[k][j])%MOD)%MOD; } } } return c;}void er_fun(int x){ tp=res; tp1=q; while (x) { if(x&1) tp=Martix_Mul(tp,tp1); tp1=Martix_Mul(tp1,tp1); x>>=1; }}int main(){ int n,i,j,sum; for(i=0;i<N;i++) { for (j=0;j<N;j++) { res.martix[i][j]=(i==j); q.martix[i][j]=0; } } q.martix[0][0]=q.martix[0][2]=q.martix[0][3]=1; q.martix[1][0]=q.martix[2][1]=q.martix[3][2]=1; elem[0]=0; elem[1]=2; elem[2]=4; elem[3]=6; elem[4]=9; while (scanf("%d%d",&n,&MOD)!=EOF) { if(n<=4) { printf("%d/n",elem[n]%MOD); continue; } sum=0; er_fun(n-4); for(i=0;i<4;i++) { sum=(sum+(tp.martix[0][i]*elem[4-i])%MOD)%MOD;if(sum<0)sum+=MOD; } printf("%d/n",sum%MOD); } return 0;}