poj 2245 Lotto

                                                                                                     Lotto

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5991		Accepted: 3806

Description

In the German Lotto you have to select 6 numbers from the set {1,2,...,49}. A popular strategy to play Lotto - although it doesn't increase your chance of winning - is to select a subset S containing k (k > 6) of these 49 numbers, and then play several games with choosing numbers only from S. For example, for k=8 and S = {1,2,3,5,8,13,21,34} there are 28 possible games: [1,2,3,5,8,13], [1,2,3,5,8,21], [1,2,3,5,8,34], [1,2,3,5,13,21], ... [3,5,8,13,21,34]. 

Your job is to write a program that reads in the number k and the set S and then prints all possible games choosing numbers only from S. 

Input

The input will contain one or more test cases. Each test case consists of one line containing several integers separated from each other by spaces. The first integer on the line will be the number k (6 < k < 13). Then k integers, specifying the set S, will follow in ascending order. Input will be terminated by a value of zero (0) for k.

Output

For each test case, print all possible games, each game on one line. The numbers of each game have to be sorted in ascending order and separated from each other by exactly one space. The games themselves have to be sorted lexicographically, that means sorted by the lowest number first, then by the second lowest and so on, as demonstrated in the sample output below. The test cases have to be separated from each other by exactly one blank line. Do not put a blank line after the last test case.

Sample Input

7 1 2 3 4 5 6 7
8 1 2 3 5 8 13 21 34
0

Sample Output

1 2 3 4 5 6 1 2 3 4 5 7 1 2 3 4 6 7 1 2 3 5 6 7 1 2 4 5 6 7 1 3 4 5 6 7 2 3 4 5 6 7 1 2 3 5 8 13 1 2 3 5 8 21 1 2 3 5 8 34 1 2 3 5 13 21 1 2 3 5 13 34 1 2 3 5 21 34 1 2 3 8 13 21 1 2 3 8 13 34 1 2 3 8 21 34 1 2 3 13 21 34 1 2 5 8 13 21 1 2 5 8 13 34 1 2 5 8 21 34 1 2 5 13 21 34 1 2 8 13 21 34 1 3 5 8 13 21 1 3 5 8 13 34 1 3 5 8 21 34 1 3 5 13 21 34 1 3 8 13 21 34 1 5 8 13 21 34 2 3 5 8 13 21 2 3 5 8 13 34 2 3 5 8 21 34 2 3 5 13 21 34 2 3 8 13 21 34 2 5 8 13 21 34 3 5 8 13 21 34

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此题是搜索第一题，DFS 虽然不懂但是琢磨了一天后还是明白了。。。。其实 return 的返回，是返回上一次调用它的地方，，，向后运行，如果函数运行到最后那么就 return；也就是说函数运行到最后的话，，，就相当于函数最后有一个return。。。。。。。

#include<stdio.h>

int f[10],a[20],n;
void DFS (int wei,int x)
{
    int i;
    if(wei==7)
    {
        for(i=1;i<=5;i++)
            printf("%d ",f[i]);
        printf("%d
",f[6]);
        return;
    }
    for(i=x+1;i<=n;i++)
    {
        f[wei]=a[i];
        DFS(wei+1,i);
    }
}

int main ()
{   
    int i,ca=1;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)break;
        if(ca!=1)printf("
");
        for(i=1;i<=n;i++)
        {
           scanf("%d",&a[i]);
        }
        for(i=1;i<=n-5;i++)
        {
            f[1]=a[i];
            DFS(2,i);
        }
        ca++;
    }
}