poj 2245 Lotto

本文介绍了一种通过编程实现德国Lotto彩票号码选择策略的方法,具体为从k个选定的号码中枚举所有可能的六数字组合。文章提供了一个完整的C++程序示例,展示了如何根据输入的k值及相应号码集合生成所有可能的游戏组合。

 

Lotto
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 6806 Accepted: 4298

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

Source

 
题意:枚举  排列组合
#include<cstdio>
using namespace std;
int s[13],K;
int main(){
    while(scanf("%d",&K)==1&&K){
        for(int i=0;i<K;i++) scanf("%d",s+i);
        for(int i=0;i+5<K;i++)
            for(int j=i+1;j+4<K;j++)
                for(int l=j+1;l+3<K;l++)
                    for(int m=l+1;m+2<K;m++)
                        for(int n=m+1;n+1<K;n++)
                            for(int o=n+1;o+0<K;o++)
                                printf("%d %d %d %d %d %d\n",s[i],s[j],s[l],s[m],s[n],s[o]);                
        printf("\n");
    }    
    return 0;
}

 

 

转载于:https://www.cnblogs.com/shenben/p/5536163.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值