hdu 1297 Children’s Queue

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 10550 Accepted Submission(s): 3392



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

计算F（n）( 转):

一：当最后一个是男孩M时候，前面n-1个随便排出来，只要符合规则就可以，即是F（n-1）；

二：当最后一个是女孩F时候，第n-1个肯定是女孩F，这时候又有两种情况：

1）前面n-2个可以按n-2个的时候的规则来，完全可以，即是F（n-2）；

2）但是即使前面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。当第n-2是女孩而n-3是男孩的情况，可能合法，情况总数为F（n-4）;

综上所述：总数F(n)=F(n-1)+F（n-2）+F(n-4);并且，F（0）=1，F(1)=1，F(2)=2，F（3）=4。
数据有点大，直接按公式计算不得行。

#include<iostream>
#include<string>
using namespace std;
string add(string a,string b)    //大数加法
{
string Max;
string Min;
if(a.size()>b.size())
{
Max=a;
Min=b;
}
else
{
Max=b;
Min=a;
}
long j=Max.size()-1;
for (long i=Min.size()-1;i>=0; i--,j--)
{
Max[j]+=Min[i]-'0';
}
for (j=Max.size()-1; j>=1; j--)
{
if(Max[j]>'9')
{
Max[j]-=10;
Max[j-1]++;
}
}
if(Max[0]>'9')
{
Max[0]-=10;
Max='1'+Max;
}
return Max;
}
int main()
{
int n;
string a[1001];
a[0]='1';
a[1]='1';
a[2]='2';
a[3]='4';
for (int i=4; i<1001; i++)
{
a[i]=add(add(a[i-1],a[i-2]),a[i-4]);
}
while (cin>>n)
{
cout<<a
<<endl;
}<span style="white-space:pre">																	</span>    return 0;
}