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POJ 2245

按字典序选一个六组合。简单DFS

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int se[20];int ans[20];void dfs(int ps,int position,int k){	if(position>k){		if(ps==6){			printf("%d",ans[1]);			for(int i=2;i<=6;i++)			printf(" %d",ans[i]);			printf("\n");		}		return ;	}	ans[ps+1]=se[position];	dfs(ps+1,position+1,k);	dfs(ps,position+1,k);}int main(){	int k;	while(scanf("%d",&k),k){		for(int i=1;i<=k;i++)		scanf("%d",&se[i]);		dfs(0,1,k);		printf("\n");	}	return 0;}

  

POJ 2245