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?

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)

For each test case,there is only one integer means the number of queue satisfied the headmaster’s needs.

1
2
3

1
2
4

SmallBeer (CML)

杭电ACM集训队训练赛（VIII）?

题目大意：主要要求是，n个人站一排，不允许女生单独站着，至少要有两个或两个以上女生咋还能在一块才合法。

思路：

1：最后一个人是男的，则只要前n-1个人的排列合法即可。

2：左后一个人是女的，则倒数第二个人必须是女的，则又分两种情况：

（1）前n-2个人的排列是合法的，则整个队列就是合法的。

（2）前n-2个不是合法的，则有这样的形式a[n-4]+男+女的形式才能使整个队列合法。

对于大数的处理，我采用的是数组，每个数组中的元素，只存10000以内的数，超过10000的在来个数组储存。

代码：

#include <iostream>
#include<cstdio>
using namespace std;
int main()
{
    int n,i,m,e,j;
    int a[1000],b[1000],c[1000],d[1000],f[1000];//定义数组，数组b,c,d,f分别存a[n-1],a[n-2],a[n-3],a[n-4];
    while(cin>>n)
    {
        a[0]=1;
        a[1]=1;
        a[2]=2;
        a[3]=4;
        if(n>=4)
        {
            b[0]=4;
            c[0]=2;
            d[0]=1;
            f[0]=1;
            m=0;
            for(i=4; i<=n; i++)
            {
                e=0;
                for(j=0; j<=m; j++)
                {
                    a[j]=b[j]+c[j]+d[j]+e;//相当于递推公式a[n]=a[n-1]+a[n-2]+a[n-4];
                    e=a[j]/10000;//判断是否超出了10000；
                    a[j]=a[j]%10000;//保证数组只存10000以内的数
                    d[j]=f[j];
                    f[j]=c[j];
                    c[j]=b[j];
                    b[j]=a[j];
                }
                if(e>0)//若超出则用新的数组来存；
                {
                    m++;
                    a[m]=e;
                    b[m]=a[m];
                    c[m]=0;
                    d[m]=0;
                    f[m]=0;
                }
            }
           //cout<<a[0]<<endl;
            printf("%d",a[m]);
           // cout<<m<<endl;
            for(i=m-1; i>=0; i--)
                printf("%04d",a[i]);
            printf("n");
        }
        else
            cout<<a[n]<<endl;
    }
    return 0;
}