Children’s Queue

学生队列问题求解

最新推荐文章于 2025-09-07 20:32:33 发布

转载最新推荐文章于 2025-09-07 20:32:33 发布 · 86 阅读

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原文链接：http://www.cnblogs.com/u013533289/p/4477323.html

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Children’s Queue

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 80 Accepted Submission(s) : 19

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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

1
2
3

Sample Output

1
2
4

#include<iostream>
#include<string> 
using namespace std;
string f(string s1,string s2)
{
    
    int j,l,p,q,m,n;
    string max,min;//将两个大数用字符串表示
    max=s1;
    min=s2;
    m=s1.length();//分别取其长度
    n=s2.length();
    if(m<n)
    {
        max=s2;
        min=s1;
    }
    p=max.size();
    q=min.size();
    l=p-1;
    for(j=q-1;j>=0;j--,l--)//从最低位开始，每一位对应相加，并且max保存和
        max[l] += min[j]-'0'; 
    for(j=p-1;j>=1;j--) //整数部分
        if(max[j]>'9')
        {
            max[j]-=10;//如果两个数字相加大于9，则进一位
            max[j-1]++;//前一位由于进位加1
        }
        if(max[0]>'9')//小数部分
        {
            max[0]-=10;
            max='1'+max;
        }
        return max;
}
int main()
{
    int n,i;
    string a[1001];
    a[0]="1";
    a[1]="1";
    a[2]="2";
    a[3]="4";
    for(i=4;i<1001;++i)
        a[i]=f(f(a[i-1],a[i-2]),a[i-4]);
        while(cin>>n)
        cout<<a[n]<<endl;
    return 0;      
}

转载于:https://www.cnblogs.com/u013533289/p/4477323.html