HDU 1297 Children’s Queue（递推）

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8915    Accepted Submission(s): 2835

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

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]。

    #include<stdio.h>
    int main(){
            int n;
            int f[1001][101] = {0};
            f[0][1] = 1;
            f[1][1] = 1;
            f[2][1] = 2;
            f[3][1] = 4;
            for(int i = 4; i < 1001; ++i){
                    for(int j = 1; j < 101; ++j){
                            f[i][j] += f[i - 1][j] + f[i - 2][j] + f[i - 4][j];     //数组的每一位相加
                            f[i][j + 1] += f[i][j] / 10000; //超过4位的部分保存至数组下一位中
                            f[i][j] %= 10000;       //每位数组只保存其中4位
                    }
            }
            while(scanf("%d", &n) != EOF){
                    int k = 100;
                    while(!f[n][k--]);      //排除前面为空的数组
                    printf("%d", f[n][k + 1]);      //输出结果的前四位
                    for(; k > 0; --k){
                            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
                    }
                    printf("
");
            }
            return 0;
    }