HDU 1297 Children’s Queue

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

思路如下：

一个长度n的队列可以看成一个n - 1的队列再追加的1个小孩，这个小孩只可能是：

a.男孩，任何n - 1的合法队列追加1个男孩必然是合法的，情况数为f[n - 1]；

b.女孩，在前n - 1的以女孩为末尾的队列后追加1位女孩也是合法的，我们可以转化为n - 2的队列中追加2位女孩；

一种情况是在n - 2的合法队列中追加2位女孩，情况数为f[n - 2]；

但我们注意到本题的难点，可能前n - 2位以女孩为末尾的不合法队列（即单纯以1位女孩结尾），也可以追加2位女孩成为合法队列，而这种n - 2不合法队列必然是由n - 4合法队列+1男孩+1女孩的结构，即情况数为f[n - 4]。

得出递推公式如下：

[code]f
 = f[n - 1] + f[n - 2] + f[n - 4]

还有需要注意的是本题中可能得出极大的数字，如当n为1000时，结果为：

[code]    12748494904808148294446671041721884239818005733501580815621713101333980596197474744336199742452912998225235910891798221541303838395943300189729514282623665199754795574309980870253213466656184865681666106508878970120168283707307150239748782319037

这种超大数完全不是C++提供的数值类型能够储存的，因此我们可以使用数组来模拟超大数运算

[code]#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const long long M = 100000+10, N = 1e9+7;
long long int n, f[1111][111];

int main()
{
#ifndef   ONLINE_JUDGE
    //freopen("1.txt", "r", stdin);
#endif
    int  i, j;
    memset(f, 0, sizeof(f));
    f[0][1] = 1;
    f[1][1] = 1;
    f[2][1] = 2;
    f[3][1] = 4;
    f[4][1] = 7;
    for (i = 5; i <= 1000; i++)
    {
        for (j = 1; j < 100; j++)
        {
            f[i][j] += f[i - 1][j] + f[i - 2][j] + f[i - 4][j];
            f[i][j+1] += f[i][j]/10000;
            f[i][j] %= 10000;
        }
    }
    while(~scanf("%d", &n))
    {
        i = 99;
        while(!f
[i--]);  
        cout << f
[i+1];
        for (; i > 0; i--)
            printf("%04d", f
[i]);
        cout << endl;
    }   
    return 0;
}