HDU - 1342 Lotto

最新推荐文章于 2019-08-22 18:19:41 发布

原创最新推荐文章于 2019-08-22 18:19:41 发布 · 300 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#ACM

ACM-深搜专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一种常见的彩票游戏玩法——从49个数字中选择6个数字的组合方式，并通过递归算法实现任意指定数量数字的所有可能组合。文章提供了一个具体的示例，即从8个数字中选择6个数字的所有可能组合，并附带了完整的C++代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

In a Lotto I have ever played, one has 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:

1,2,3,5,8,131,2,3,5,8,13,

1,2,3,5,8,211,2,3,5,8,21,

1,2,3,5,8,341,2,3,5,8,34,

1,2,3,5,13,211,2,3,5,13,21, ...

3,5,8,13,21,343,5,8,13,21,34.

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 file 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

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,a[15];
int vis[15];
void DFS(int num,int k) {
    int i;
    if(k==6) {
        int ii=0;
        for(i=0; i<n; i++) {
            if(vis[i]) {
                ii++;
                if(ii==6)
                    printf("%d\n",a[i]);
                else
                    printf("%d ",a[i]);
            }
        }
        return ;
    }
    for(i=num+1; i<n; i++) {
        vis[i]=1;
        DFS(i,k+1);
        vis[i]=0;
    }
}
int main() {
    int i,l;
    int kk=0;
    while(scanf("%d",&n) && n) {
            if(kk!=0)
            printf("\n");
        for(i=0; i<n; i++)
            scanf("%d",&a[i]);
        for(i=0; i<=n-6; i++) {
            memset(vis,0,sizeof(vis));
            vis[i]=1;
            DFS(i,1);
        }
    kk++;
    }
}