Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9549 Accepted Submission(s): 3053
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
Sample Output
解题思路：
一个长度n的队列可以看成一个n - 1的队列再追加的1个小孩，这个小孩只可能是：
a.男孩，任何n - 1的合法队列追加1个男孩必然是合法的，情况数为f[n - 1]；
b.女孩，在前n - 1的以女孩为末尾的队列后追加1位女孩也是合法的，我们可以转化为n - 2的队列中追加2位女孩；
一种情况是在n - 2的合法队列中追加2位女孩，情况数为f[n - 2]；
但我们注意到本题的难点，可能前n - 2位以女孩为末尾的不合法队列（即单纯以1位女孩结尾），也可以追加2位女孩成为合法队列，而这种n - 2不合法队列必然是由n - 4合法队列+1男孩+1女孩的结构，即情况数为f[n - 4]。
得出递推公式：f[n]=f[n-1]+f[n-2]+f[n-4];
代码如下：
#include"stdio.h"
#include"string.h"
#define N 1000
#define M 10000
int main()
{
int i,j,n,len=100;
int f[N][100];
memset(f,0,sizeof(f));
f[0][0]=1;
f[1][0]=2;
f[2][0]=4;
f[3][0]=7;
for(i=4;i<N;i++)
{
for(j=0;j<len;j++)
{
f[i][j]+=f[i-1][j]+f[i-2][j]+f[i-4][j];
f[i][j+1]+=f[i][j]/M;
f[i][j]%=M;
}
}
while(scanf("%d",&n)!=-1)
{
i=len-1;
while(!f[n-1][i--]); //这个很不错啊!（简洁）
printf("%d",f[n-1][i+1]);
for(;i>=0;i--)
printf("%04d",f[n-1][i]);
puts("");
}
return 0;
}
hdu 1297 Children’s Queue
原文：http://blog.csdn.net/u011721440/article/details/19344453
