(hdu step 3.1.7)Children’s Queue(求n个人站在一起有m个人必须是连在一起的方案数)

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题目：

Children’s Queue

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

Total Submission(s): 853 Accepted Submission(s): 479

[align=left]Problem Description[/align]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?

[align=left]Input[/align]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)

[align=left]Output[/align]For each test case, there is only one integer means the number of queue satisfied the headmaster’s needs.

[align=left]Sample Input[/align] 1 2 3

[align=left]Sample Output[/align] 1 2 4

[align=left]Author[/align]SmallBeer (CML)

[align=left]Source[/align]杭电ACM集训队训练赛（VIII）

[align=left]Recommend[/align]lcy

题目分析：
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]。因为这道题的n已经达到1000了，而且又是“累加”这种计算模型，所以果断使用java的大数来做

代码如下：
import java.math.BigInteger;
import java.util.Scanner;

public class Main {

public final static int maxn = 1001;
static BigInteger dp[] = new BigInteger[maxn];

public static void prepare(){
dp[1] = BigInteger.valueOf(1);
dp[2] = BigInteger.valueOf(2);
dp[3] = BigInteger.valueOf(4);
dp[4] = BigInteger.valueOf(7);

int i;
for(i = 5 ; i < maxn ; ++i){
dp[i] = new BigInteger("0");
dp[i] = dp[i].add( dp[i-1]).add( dp[i-2]).add( dp[i-4]);
}
}

public static void main(String[] args) {
prepare();

Scanner scanner = new Scanner(System.in);
while(scanner.hasNext()){
int n = scanner.nextInt();

System.out.println(dp
);
}
}
}