hdu 1297

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 959 Accepted Submission(s): 534

Problem Description

There are many students in PHT School. One day, the headmaster whose name is PigHeader wanted all students stand in a line. He prescribed that girl can not be in single. In other words, either no girl in the queue or more than one girl stands side by side. The case n=4 (n is the number of children) is like
FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM
Here F stands for a girl and M stands for a boy. The total number of queue satisfied the headmaster’s needs is 7. Can you make a program to find the total number of queue with n children?

 

Input

There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer n means the number of children (1<=n<=1000)

 

Output

For each test case, there is only one integer means the number of queue satisfied the headmaster’s needs.

 

Sample Input

1
2
3

 

Sample Output

1
2
4

 大数处理加dP,,dp的话f[n]=f[n-1]+f[n-2]+f[n-4]

假设当前要加入第n个学生，如果他是男生，则加入对前面的序列都不会构成影响，

即为f[n-1]，假如他是女生，为保证合法，她前面的那个也就是第n-1个也必须是

女生，接下来分两种情况讨论，如果第n-2个及以前的都合法，那么显然加入最后

这两个女生也是合法的额，但是如果不合法呢？很明显，如果不合法，问题应该出在

尾部，也就是原来f[n-2]不合法，但是加入最后两个女生后也就变得合法了（这句话

很关键），那么其只能为f[n-4]+男+女，所以即可得到公式

其实这个讨论的关键在于确定最后两个为女生才能保证合法

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
long long dp[1001][1001];
int main()
{
   int n;
   memset(dp,0,sizeof(dp));
   dp[1][0]=1;dp[1][1]=1;
   dp[2][0]=1;dp[2][1]=2;
   dp[3][0]=1;dp[3][1]=4;
   dp[4][0]=1;dp[4][1]=7;
   for(int i=5;i<=1000;i++)
     {
       int j,t=dp[i-1][0];
       for(j=1;j<=t;j++)
        {
           dp[i][j]+=dp[i-1][j]+dp[i-2][j]+dp[i-4][j];
           if(dp[i][j]>=10)
           {
               dp[i][j+1]+=dp[i][j]/10;
               dp[i][j]%=10;
           }
        }
       while(dp[i][j]>=10)
       {
           dp[i][j+1]+=dp[i][j]/10;
           dp[i][j]%=10;
           j++;
       }
       j=1000;
       while(!dp[i][j])
           j--;
       dp[i][0]=j;
     }
   while(cin>>n)
   {
      for(int i=dp[n][0];i>=1;i--)
         cout<<dp[n][i];
      cout<<endl;
   }
   return 0;
}