yt13递推Children’s Queue （大数）

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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

Sample Output

Author

SmallBeer (CML)

Source

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;}

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