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; }

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 c; cin>>c; if (c >= 'A' && c <= 'Z') { cout << "A" << endl; } else if (c >= 'a' && c <= 'z') { cout << "a" << endl; } else if (c >= '0' && c <= '9') { cout << "0" << endl; } else { cout << "other" << endl; } }

#include <bits/stdc++.h> using namespace std; int main() { char a; cin >> a; if(int(a)>=65 && int(a)<=90) { cout << "A"; }else if(int(a)>=97 && int(a)<=122) { cout << "a"; }else if(int(a)>=48 && int(a)<=57) { 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 c; cin >> c; // 读取一个字符 }

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#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; }

也是非常的简单