Children’s Queue 1297_children鈥檚 queuecsdn-优快云博客

本文介绍了一个有趣的算法问题——学生排队问题。问题要求女孩不能单独站在队伍中，并给出了一个递推公式来解决该问题，通过预计算的方式快速得出任意数量学生可能的排队组合总数。

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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

#include <iostream>
#include <string>

std::string add(std::string strA, std::string strB);

int main(int argc, const char *argv[])
{
    std::string strResult[1000] = {"1", "2", "4", "7"};
    for(int i = 4;i < 1000;++ i)
    {
        strResult[i] = add(add(strResult[i - 1], strResult[i - 2]), strResult[i - 4]);
    }

    int n = 0;
    while(std::cin >> n)
    {
        std::cout << strResult[n - 1] << std::endl;
    }

    //system("pause");
    return 0;
}

std::string add(std::string strA, std::string strB)
{
    int nLenMax, nLenMin;
    std::string strMax, strMin;
    if(strA.size() >= strB.size())
    {
        nLenMax = strA.size();
        strMax = strA;
        nLenMin = strB.size();
        strMin = strB;
    }
    else
    {
        nLenMax = strB.size();
        strMax = strB;
        nLenMin = strA.size();
        strMin = strA;
    }

    for(int i = nLenMin - 1, j = nLenMax - 1;i >= 0;-- i, -- j)
    {
        strMax[j] += strMin[i] - '0';
    }
    for(int i = nLenMax - 1;i > 0;-- i)
    {
        if(strMax[i] > '9')
        {
            strMax[i] -= 10;
            ++ strMax[i - 1];
        }
    }
    if(strMax[0] > '9')
    {
        strMax[0] -= 10;
        strMax = '1' + strMax;
    }

    return strMax;
}