L=[] num=input() num=num:-1 for i in num: s=i+' ' L.append(s) num="".join(L) print(num

#include <iostream> using namespace std; int main() { int n,q,w,e; cin>>n; q=n/100; w=n/10%10; e=n%10; cout<<e<<" "<<w<<" "<<q<<endl; return 0; }

#include <bits/stdc++.h> using namespace std; int main(){ int n,a,b,c,d,e; cin>>n; c=n%10; d=n-c; b=d/10%10; e=d-10*b; a=d/100; cout<<c<<" "<<b<<" "<<a; return 0; }

#include <bits/stdc++.h> using namespace std; int main(){ int a; cin>>a; while(a>0){ int b=a%10; a/=10; cout<<b<<" "; } return 0; }

第二题解 通过分位截取三位数的字符串形态并倒叙输出 (不是最优解，但对于普通写法【% /】有了巨大提升)

#include<iostream> using namespace std; void trouble(){ int a; cin>>a; cout<<a%10<<" "<<a/10%10<<" "<<a/100; } int main(){ trouble(); }

#include<iostream> using namespace std; string a; int main(){ cin>>a; cout<<a[2]<<' '<<a[1]<<' '<<a[0]; return 0; }

#include<bits/stdc++.h> using namespace std; int main(){ int n; cin>>n; cout<<n%10<<" "<<n/10%10<<" "<<n/100; return 0; }

char的解法

#include<iostream> using namespace std; int a,b,c,d; int main(){ cin>>a; b=a/100; c=a/10%10; d=a%10; cout<<d<<" "<<c<<" "<<b; return 0; }

#include<iostream> using namespace std; int main(){ int a; cin>>a; int ge = a%10,shi = a/10%10,bai = a/100; cout<<ge<<" "<<shi<<" "<<bai; return 0; }

秒啦！！！

#include <bits/stdc++.h> using namespace std; int main(){ int a; cin>>a; while(a>0){ int b=a%10; a/=10; cout<<b<<" "; } return 0; }

#include <bits/stdc++.h> using namespace std; int main(){ int a; cin>>a; cout <<a%10<<" "<<a%100/10<<" "<<a/100; }

#include <bits/stdc++.h> using namespace std; int main(){ int n; cin>>n; while(n>0){ int m=n%10; n/=10; cout <<m<<" "; } return 0; }

最爱string偷懒了 #include <bits/stdc++.h> using namespace std; int main() { string s; cin>>s; for(int i=s.size()-1;i>0;i--){ cout<<s[i]<<" "; }cout<<s[0]; return 0; }