Queuing

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2842 Accepted Submission(s): 1305

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.

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Sample Output

6
2
1

Author WhereIsHeroFrom
递推公式f(n)=f(n-1)+f(n-3)+f(n-4) 递推过程： 现在设长度为n 1，在最后一位加m时，前面无论是什么都满足，得f(n-1); 2，在 最后一位加f时，倒数后2位无论加什么都不能满足，于是讨论倒数后3位，后3位可以加mmf,现在还有一种情况没考虑到，那就是mff，可能前面是以f结尾，于是讨论倒数后4位，在后4位上加mmff。 得到递推公式就可以构造矩阵了。

代码：

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       using namespace std; struct matrix { int ma[10][10]; }; int mod; matrix multi(matrix x,matrix y) { matrix ans; memset(ans.ma,0,sizeof(ans.ma)); for(int i=1;i<=4;i++) { for(int j=1;j<=4;j++) { if(x.ma[i][j]) { for(int k=1;k<=4;k++) { ans.ma[i][k]=(ans.ma[i][k]+(x.ma[i][j]*y.ma[j][k])%mod)%mod; } } } } return ans; } matrix pow(matrix a,int k) { matrix ans; for(int i=1;i<=4;i++) { for(int j=1;j<=4;j++) { if(i==j) ans.ma[i][j]=1; else ans.ma[i][j]=0; } } while(k) { if(k&1) ans=multi(ans,a); a=multi(a,a); k=k>>1; } return ans; } int main() { int n; while(~scanf("%d%d",&n,&mod)) { matrix a,b; memset(a.ma,0,sizeof(a.ma)); memset(b.ma,0,sizeof(b.ma)); a.ma[1][1]=a.ma[1][3]=a.ma[1][4]=1; a.ma[2][1]=a.ma[3][2]=a.ma[4][3]=1; b.ma[1][1]=6; b.ma[2][1]=4; b.ma[3][1]=2; b.ma[4][1]=1; if(n==0) { printf("%d\n",1%mod); continue; } if(n==1) { printf("%d\n",2%mod); continue; } if(n==2) { printf("%d\n",4%mod); continue; } if(n==3) { printf("%d\n",6%mod); continue; } a=pow(a,n-3); b=multi(a,b); printf("%d\n",b.ma[1][1]); } return 0; }

hdu 2604 Queuing（矩阵优化递推公式）

Queuing