Children's Queue (recursive)

Children's Queue (recursive)
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 missing 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

123

Sample Output

124

Question: There are n locations where boys and girls are waiting in line and at least two girls are required to be together. Thought: set f [n] to indicate the situation of n people. Case 1: Add a boy after f [n-1]; Case 2: Add two girls after f [N-2; case 3: add three girls to the end of f [n-3] (equivalent to f [n-4] Number); so: f [n] = f [n-1] + f [N-2] + f [n-4]; because of the large data, so the use of large number addition can be. Reprinted please indicate the source: Find and star child questions link: http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 1297

#include
 
  #include
  
   int f[1005][105];void init(){    memset(f,0,sizeof(f));    f[0][1]=1;    f[1][1]=1;    f[2][1]=2;    f[3][1]=4;    for(int i=4;i<=1000;i++)    {        int add=0;        for(int j=1;j<=100;j++)        {            f[i][j]=f[i-1][j]+f[i-2][j]+f[i-4][j]+add;            add=f[i][j]/10000;            f[i][j]%=10000;            if(add==0&&f[i][j]==0)break;        }    }}int main(){    int n;    init();    while(scanf("%d",&n)!=EOF)    {        int k=100;        while(!f[n][k])k--;        printf("%d",f[n][k--]);        for(;k>0;k--)        {            printf("%04d",f[n][k]);        }        printf("\n");    }    return 0;}