2171: An Easy Problem

 2171: An Easy Problem!

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Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	3s	8192K	627	186	Standard

It's an easy problem!

N-bit sequences is a string of digitals which contains only '0' and '1'. You should determine the number of n-bit sequences that contain no three continuous 1's. For example, for n = 3 the answer is 7 (sequences 000, 001, 010, 011, 100, 101, 110 are acceptable while 111 is not).

Input

For each line, you are given a single positive integer N no more than 40 on a line by itself.

Output

Print a single line containing the number of n-bit sequences which have no three continuous 1's.

Sample Input

1
2
3

Sample Output

2
4
7

动态规划。所有的串都能按照最末两位分成四组。假设在n位合法的0、1串中，以00结尾的串的个数为f0(n)，以01结尾的串个数为f1(n)，以10结尾的串个数为f2(n)，以11结尾的串个数为f3(n)，则n位合法的0、1串的总数f(n)=f0(n)+f1(n)+f2(n)+f3(n)。当串的长度扩展到n+1位时，我们看这它们各自的个数如何动态变化。从n位到n+1位，只需要在n位串的末尾追加一位0或一位1即可。对四个函数分类讨论：
以00结尾：加0后变成以00结尾，加1后变成01结尾；
以01结尾：加0后变成以10结尾，加1后变成11结尾；
以10结尾：加0后变成以00结尾，加1后变成01结尾；
以11结尾：加0后变成以10结尾，加1后不合法。
所以，有：　f0(n+1) = f0(n) + f2(n),  　f1(n+1) = f0(n) + f2(n),  f2(n+1) = f1(n) + f3(n), 　f3(n+1) = f1(n)。
当n为2时，有f0(2) = f1(2) = f2(2) = f3(2) = 1。这样，经过递推可以很快计算出结果。中间要考虑溢出的情况。当n = 37时，结果超过了70亿，必须用double类型。

整理得到  a[i]=a[i-1]+a[i-2]+a[i-3];

#include<stdio.h>
int main()
{
    double a[50]={0,2,4,7};
    for( int i=4;i<=40;i++)
    {
        a[i]=a[i-1]+a[i-2]+a[i-3];
    }
    int n;
    while(scanf("%d",&n)==1)
    {
        printf("%.0lf/n",a[n]);
    }
    return 0;
}