Children’s Queue（hdu1297+递推）

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12101 Accepted Submission(s):
3953

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

 

题意：有n个位置，男孩女孩排队，要求女孩至少要2个在一起。

 

思路：设f[n]表示，n个人的情况。情况一、在f[n-1]的情况后面加一个男孩；情况二、在f[n-2]的情况后面加两个女孩；情况三、在f[n-3]最后是男孩（等价于在f[n-4]个个数）的后面加三个女孩；

所以：f[n]=f[n-1]+f[n-2]+f[n-4];由于数据比较大，所以采用大数加法就可以了。

 

转载请注明出处：寻找&星空の孩子 题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1297

 

 #include<stdio.h>
 #include<string.h>
 int f[][];
 void init()
 {
     memset(f,,sizeof(f));
     f[][]=;
     f[][]=;
     f[][]=;
     f[][]=;
 
     for(int i=;i<=;i++)
     {
         int add=;
         for(int j=;j<=;j++)
         {
             f[i][j]=f[i-][j]+f[i-][j]+f[i-][j]+add;
             add=f[i][j]/;
             f[i][j]%=;
             if(add==&&f[i][j]==)break;
         }
     }
 }
 int main()
 {
     int n;
     init();
     while(scanf("%d",&n)!=EOF)
     {
         int k=;
         while(!f[n][k])k--;
         printf("%d",f[n][k--]);
         for(;k>;k--)
         {
             printf("%04d",f[n][k]);
         }
         printf("\n");
     }
     return ;
 }