Children’s Queue

Problem  Link：http://acm.hdu.edu.cn/showproblem.php?pid=1297

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10591    Accepted Submission(s): 3407

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
  
  

 

Author

SmallBeer (CML)

 

Source

杭电ACM集训队训练赛（VIII）

code：

#include<stdio.h>
int f[1001][246];
void init()
{
int i,j,k,g;
k=1;
f[1][1]=1;f[1][0]=1;//f[i][0]表示第i个数位数
f[2][1]=2;f[2][0]=1;
f[3][1]=4;f[3][0]=1;
f[4][1]=7;f[4][0]=1;
for(i=5;i<=1000;i++)
{
g=0;
for(j=1;j<=k;j++)
{
 f[i][j]=f[i-1][j]+f[i-2][j]+f[i-4][j]+g;
 g=f[i][j]/10;
 f[i][j]%=10;
 f[i][j+1]=g;
}
while(f[i][j]>0)
{
k++;
f[i][j+1]=f[i][j]/10;
f[i][j]%=10;
j++;
}
f[i][0]=k;
}
}
int main()
{
int n,i;
init();
//printf("%d\n",f[1000][0]);==245 先开较大的数组在输出最大的位数 调整
while(scanf("%d",&n)!=EOF)
{
for(i=f[n][0];i>=1;i--)
printf("%d",f[n][i]);
printf("\n");
}
return 0;
}