hdu-1297（找规律+大数加法）

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others) ? ?Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10643 ? ?Accepted Submission(s): 3430

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

思路：分男女生考慮，設總共有n個人(f(n)表示當有n個人的時候合法序列的個數，ak表示第k個人)，當a1為男生時，有f(n-1)種可能，當a1是女生時，a2必為女生；當a1、a2都為女生時，a3為男生，有f(n-3)種可能，當a1、a2、a3都為女生，a4為男生時，有f(n-4)種可能，以此類推（下式中1表示n全為女生時那種情況，規定f(0) = 1）,
故有 f(n) = f(n-1) + f(n-3) + f(n-4) + f(n-5) + ... + f(1) + f(0) + 1;
所以 f(n-2) = ? ? ? f(n-3) + ? ? ? ? f(n-5) + f(n-6) + ... + f(1) + f(0) + 1;

由以上兩式得 f(n) = f(n-1) + f(n-2) + f(n-4)

#include<iostream>
#include<string>
#include<stdio.h>
using namespace std;
string fs(string a,string b)
{
	string c;
	int la=a.length()-1;
	int lb=b.length()-1;
	int k=0;
	while(la>=0||lb>=0)
	{
		if(la>=0&&lb>=0){
			c=char((a[la]+b[lb]-96+k)%10+48)+c;
			k=(a[la]+b[lb]-96+k)/10;
			la--;lb--;
		}
		else if(la>=0){
			c=char((a[la]-48+k)%10+48)+c;
			k=(a[la]+k-48)/10;
			la--;
		}
		else if(la>=0){
			c=char((b[lb]-48+k)%10+48)+c;
			k=(b[lb]+k-48)/10;
			lb--;
		}
	}
	if(k!=0) c=char(k+48)+c;
	return c;
}
string f[1001];
int main()
{
	int n,m;
	f[1]="1";
	f[2]="2";
	f[3]="4";
	f[4]="7";
	for(int i=5;i<=1000;i++)
	{
		string y;
		y=fs(f[i-1],f[i-2]);
		f[i]=fs(y,f[i-4]);
	}
	while(scanf("%d",&n)!=EOF)
		cout<<f[n]<<endl;
	return 0;
}