题解(稍微有难度，附详细注释): 这道题考察的是我们对ASCII码表的熟悉度，一起来看看正解吧~ qwq写注释真的很累，希望大家点上一个大大的赞，这是我更新的动力~

#include<bits/stdc++.h> using namespace std; int main(){ char a; cin>>a; if(int(a)>=97&&int(a)<='z'){ }

题解

#include<iostream> using namespace std; int main(){ char a; cin>>a; if(a>=97 && a<=122){ cout<<"a"; }else if(a>=65 && a<=90){ cout<<"A"; }else if(a>=48 && a<=57){ cout<<"0"; }else{ cout<<"other"; } return 0; }

也是非常的简单

简单！

其实这道题不用ASCII，不过是判断在哪两个字符之间罢了。只要结合if语句，这道题并不难。 #include<bits/stdc++.h> using namespace std; int main(){ char n; cin>>n; if(n>='a' and n<='z') cout<<'a'; else if(n>='A' and n<='Z') cout<<'A'; else if(n>='0' and n<='9') cout<<0; else cout<<"other"; return 0; }

#include<bits/stdc++.h> using namespace std; char a; int main(){ cin>>a; if(a>='A'&&a<='Z')cout<<"A"; else if(a>='a'&&a<='z')cout<<'a'; else if(a>='1'&&a<='9')cout<<"0"; else cout<<"other"; return 0; }

#include<bits/stdc++.h> using namespace std; int main(){ char a; cin>>a; if(a>='A'&&a<='Z'){ cout<<'A'; }else if(a>='a'&&a<='z'){ cout<<'a'; }else if(a>='0'&&a<='9'){ cout<<'0'; }else{ cout<<"other"; } return 0; }

#include <iostream> using namespace std; int main(){ char n; cin>>n; if(n>=65 and n <= 90){ cout<<'A'; } else if(n>=97 and n<=122){ cout<<'a'; } else if(n>=48 and n<=57){ cout<<'0'; } else{ cout<<"other"; } return 0; }

防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视防透视

jhtgurtfyt

#include<iostream> using namespace std; int main(){ char a; cin>>a; if(a>='a'&&a<='z'){ cout<<"a"<<endl; }else if(a>='A'&&a<='Z'){ cout<<"A"<<endl; }else if(a>='0'&&a<='9'){ cout<<0<<endl; }else{ cout<<"other"; } }