hdoj 1342 Lotto【dfs】
Lotto
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1736 Accepted Submission(s):
854
Problem Description
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.
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
做完搜索真的感觉自己的智商太不够用了....
#include<stdio.h>
#include<string.h>
int a[50];
int b[10];
int t;
void dfs(int x,int y)
{
int i;
if(x==7)
{
for(i=1;i<=6;i++)
{
if(i==1)
printf("%d",b[i]);
else
printf(" %d",b[i]);
}
printf("\n");
return ;
}
if(y>t) return ;
b[x]=a[y];
x++;
dfs(x,y+1);
x--;
dfs(x,y+1);
}
int main()
{
int n,m,j,i;
n=0;
while(scanf("%d",&t),t)
{
if(n)
printf("\n");
for(i=1;i<=t;i++)
scanf("%d",&a[i]);
dfs(1,1);
n++;
}
return 0;
}
#include<stdio.h>
#include<string.h>
int n,m;
int a[30],b[30];
void dfs(int cur,int pos)
{
int i,j;
if(pos==7)
{
for(i=1;i<7;i++)
{
if(i==1) printf("%d",b[i]);
else printf(" %d",b[i]);
}
printf("\n");
return ;
}
if(cur>n) return ;
b[pos]=a[cur];//将a数组的值给b数组
dfs(cur+1,pos+1);//a数组和b数组同时自身加1移向下一位
dfs(cur+1,pos);//如果上边搜索失败回溯回来则令b数组从后向前改变值
}
int main()
{
int i,j;
int t=0;
while(scanf("%d",&n),n)
{
if(t)
printf("\n");
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
dfs(1,1);
t++;
}
return 0;
}
