Children’s Queue .

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5660 Accepted Submission(s): 1755

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

Source

HDU1297

思路如下：

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]。

若感觉本题较难，可先查看文章：[ACM_HDU_2045]LELE的RPG难题，思路与本题类似，但较为简单。

得出递推公式如下：

f[n] = f[n - 1] + f[n - 2] + f[n - 4]

还有需要注意的是本题中可能得出极大的数字，如当n为1000时，结果为：

12748494904808148294446671041721884239818005733501580815621713101333980596197474744336199742452912998225235910891798221541303838395943300189729514282623665199754795574309980870253213466656184865681666106508878970120168283707307150239748782319037

这种超大数完全不是C++提供的数值类型能够储存的，因此我们可以使用二维数组来模拟超大数运算，代码如下：

 

01.#include<stdio.h>   02.int main(){  03.    int n;  04.    int f[1001][101] = {0};  05.    f[0][1] = 1;  06.    f[1][1] = 1;  07.    f[2][1] = 2;  08.    f[3][1] = 4;  09.    for(int i = 4; i < 1001; ++i){  10.        for(int j = 1; j < 101; ++j){  11.            f[i][j] += f[i - 1][j] + f[i - 2][j] + f[i - 4][j]; //数组的每一位相加   12.            f[i][j + 1] += f[i][j] / 10000; //超过4位的部分保存至数组下一位中   13.            f[i][j] %= 10000;   //每位数组只保存其中4位   14.        }  15.    }  16.    while(scanf("%d", &n) != EOF){  17.        int k = 100;  18.        while(!f[n][k--]);  //排除前面为空的数组   19.        printf("%d", f[n][k + 1]);  //输出结果的前四位   20.        for(; k > 0; --k){  21.            printf("%04d", f[n][k]);    //输出其余的所有四位数字，若数字小于四位，则前面用0填充   22.        }  23.        printf("\n");  24.    }  25.    return 0;  26.}