#include<iostream>using namespace std;int source[]={1,2,3,4,5,6,7,8,9},n=9,m=3,record[10],visited[10],counter=0;void output(){for(int i=0;i<m;i++)cout<<record[i]<<" ";cout<<endl;}void dfs(int step){if(step==m){output();counter++;return ;}for(int i=0;i<n;i++){if(visited[i]==0){visited[i]=1;record[step]=source[i];dfs(step+1);visited[i]=0;}}}int main(){dfs(0);cout<<counter<<endl;return 0;}
,而做人的能力则会给你一百种机会。