HDUOJ 1297

原题：

• 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

解题思路：

• 按照最后一个人的性别分析，他要么是男，要么是女，所以可以分两大类讨论：
  
  • 如果n个人的合法队列的最后一个人是男，则对前面n-1个人的队列没有任何限制，他只要站在最后即可，所以，这种情况一共有F(n-1);
  • 如果n个人的合法队列的最后一个人是女，则要求队列的第n-1个人务必也是女生，这就是说，限定了最后两个人必须都是女生，这又可以分两种情况：
    
    • 如果队列的前n-2个人是合法的队列，则显然后面再加两个女生，也一定是合法的，这种情况有F(n-2);
    • 但是，难点在于，即使前面n-2个人不是合法的队列，加上两个女生也有可能是合法的，当然，这种长度为n-2的不合法队列，不合法的地方必须是尾巴，就是说，这里说的长度是n-2的不合法串的形式必须是“F(n-4)+男+女”，这种情况一共有F(n-4).
• 所以，通过以上的分析，可以得到递推的通项公式：
  
  • F(n)=F(n-1)+F(n-2)+F(n-4) (n>3)
• 然后就是对n<=3 的一些特殊情况的处理了
  
  • 显然：F(0)=1 (没有人也是合法的，这个可以特殊处理，就像0的阶乘定义为1一样)
  • F(1)=1
  • F(3)=4

代码：

递推求解–找递推公式

#include<stdio.h>#include <string>#include <iostream>using namespace std;string dp[1001] = { "1","1","2","4" };string BigNumAdd(string a, string b){    int A_Len = a.length() - 1, B_Len = b.length() - 1, i, temp, CarryBit = 0;    //前面补0，弄成长度相同    if (A_Len<B_Len)    {        for (i = 1;i <= B_Len - A_Len;i++)            a = "0" + a;    }    else    {        for (i = 1;i <= A_Len - B_Len;i++)            b = "0" + b;    }    //------    for (i = A_Len; i >= 0; i--)    {        temp = a[i] - '0' + (b[i] - '0') + CarryBit;        CarryBit = temp / 10;//进位        temp %= 10;        a[i] = char(temp + '0');    }    if (CarryBit != 0)a = char(CarryBit + '0') + a;    return a;}int main(){    int  n, i;    for (i = 4; i <= 1000; i++)        dp[i] = BigNumAdd(BigNumAdd(dp[i - 1], dp[i - 2]), dp[i - 4]);    while (scanf("%d", &n) != EOF)        cout << dp[n] << endl;}