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

简单！

#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; }

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; }

jhtgurtfyt

#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; }

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

#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"; } }

*** {#include<iostream> using namespace std; int main(){ }

太简单了

using namespace std; int main() { char ch; cin >> ch; }

#include<bits/stdc++.h> using namespace std; int main(){ char n; cin>>n; if(n>=65&&n=<90){ cout<<'A'; } else if(n>=97&&n=<122){ cout<<'a'; } else if(n>=48&&n=<57){ cout<<'0'; } else{ cout<<"other"; } }