JOJ 1182: Lotto _lotto程序设计-优快云博客

本文探讨了一种常见的彩票游戏玩法，即从49个数字中选择6个数字进行组合的问题。通过递归算法实现了从给定的更大集合中挑选特定数量数字的所有可能组合，并详细介绍了输入输出规范及样例。

深度搜索，递归而已

Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	1s	8192K	531	280	Standard

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: [1,2,3,5,8,13], [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 Specification

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 Specification

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<stdio.h>
int f[6],a[20],n;
void dfs(int depth,int x)
{
int i;
if(depth==6)
{
  for(i=0;i<5;i++) printf("%d ",f[i]);
  printf("%d/n",f[5]);
  return;
}
for(i=x+1;i<n;i++)
{
  f[depth]=a[i];
  dfs(depth+1,i);
}
}
int main()
{
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
int m=0,i;
while((scanf("%d",&n),n)!=0)
{
  if(m!=0)   printf("/n");
  for(i=0;i<n;i++) scanf("%d",&a[i]);
  for(i=0;i<=n-6;i++)
  {
   f[0]=a[i];
   dfs(1,i);
  }
  m=1;
}
return 0;
}