• 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]。

     1     #include<stdio.h>
     2     int main(){
     3             int n;
     4             int f[1001][101] = {0};
     5             f[0][1] = 1;
     6             f[1][1] = 1;
     7             f[2][1] = 2;
     8             f[3][1] = 4;
     9             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("
    ");
    24             }
    25             return 0;
    26     }
    View Code
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  • 原文地址:https://www.cnblogs.com/qiu520/p/3257543.html
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