#include<bits/stdc++.h>
using namespace std;
char a[1000000];
int main(){
	cin>>a;
	for(int i=strlen(a)-1;i>=0;i--){
		cout<<a[i];
	}
} 

GCC

#include<stdio.h>
char a , b , c;
int main(){
    scanf("%c%c%c" , &a , &b , &c);
    printf("%c%c%c\n" , c , b , a);
    return 0;
}

#include<bits/stdc++.h> using namespace std; char a[1000000]; int main(){ cin>>a; for(int i=strlen(a)-1;i>=0;i--){ cout<<a[i]; } }

#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
char a[1000000];
int main()
{
	cin >> a;
	for ( int i = strlen(a) - 1 ; i >= 0 ; i-- )
	{
		cout << a[i];
	}
} 
//菜鸟驿站
//老六专属

#include <bits/stdc++.h>
using namespace std;

int main()
{
    string n;
    cin >> n;
    cout << n[2] << n[1] << n[0] << endl;
    return 0;
}

#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
	int n;
	cin >> n;
	int a,b,c;
	c = n % 10;
	a = n / 100;
	b = n / 10 % 10;
	cout << c << b << a << endl;
}

#include<bits/stdc++.h>
#include<cstring>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define ll long long
using namespace std;
const int N=1e5+10;
const int M=2023;
const int inf=0x3f3f3f3f;
int n,m,k;
void dfs(int n)
{
	if(n<=0)return;
	cout<<n%10;
	dfs(n/10);
}
int main()
{
	cin>>n;
	dfs(n);
	return 0;
}

#include<iostream>
using namespace std;
//#define LL long long
//const int N = 1e6 + 10;
//const int INF = 0x3f3f3f3f;
//int a[10][10];
int a;
int main(){
	cin>>a;
	while(a){
		cout<<a%10;
		a/=10;
	}
}

#include 
using namespace std;
int main() 
{
    string n;
    getline(cin, n);
    for (int i = n.size() - 1; i >= 0; i--) 
	{
        cout << n[i];
    }
    return 0;
}

这题很简单，建议用字符串做，不然那个“0”会很麻烦(>ω･* )ﾉ

#include<iostream>
using namespace std;
int main(){
	string a;
	cin>>a;
	for(int i = a.size()-1; i >= 0; i -- ){
		cout<<a[i];
	}
	return 0;
}

#include<stdio.h> #include using namespace std; int main() { char a,b,c; cin >>a >>b >>c; cout<< c<< b<< a;

}

#include <bits/stdc++.h> using namespace std; int main() { int a; cin >> a; int b = a % 10; int c = a-b; int c2 = c % 100; int d = a - c2 - b; int d2 = d / 100; int b2 = b * 100; int e = b2 + c2 + d2; cout << e; return 0; }

#include <bits/stdc++.h> using namespace std; int main() { int a; cin >> a; int b = a % 10; int c = a-b; int c2 = c % 100; int d = a - c2 - b; int d2 = d / 100; int b2 = b * 100; int e = b2 + c2 + d2; cout << e; return 0; }

递归：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<iomanip>
#include<queue>
#include<algorithm>
#include<vector>
#include<stack>
#include<set>
using namespace std;
#define LL long long
const int N=100;
using namespace std;
int f(int n,int m)
{
	f(n%2)
}
int main()
{
	int a,b;
	cin>>a>>b;
	cout<<f(a,b);
	return 0;
}

#include<stdio.h>
int main()
{
    int n,a,b,c;
    scanf("%d",&n);
    a=n/100;
    b=n/10%10;
    c=n%10;
    printf("%d%d%d\n",c,b,a);
    return 0;
}

#include using namespace std; int main(){ int n; cin>>n; cout<<n%10; cout<<n%100/10; cout<<n/100; return 0;

}

#include

using namespace std;

int a[1111];

int main(){ int n;

cin>>n;
for(int i=1;i<=3;i++){
	cout<<n%10;
	n/=10;
}

}

需要这么难？

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
const int N=1e7+10;
const int INF=0x3f3f3f3f;
int main()
{
	int a;
	cin>>a;
	cout<<a%10<<a/10%10<<a/100%10;
}

#include using namespace std; int main(){ int n; cin>>n; cout<<n%10; cout<<n%100/10; cout<<n/100; return 0;

}

#include<iostream>
using namespace std;
int main(){
	string a;
	cin>>a;
	for(int i = a.size()-1; i >= 0; i -- ){
		cout<<a[i];
	}
	return 0;
}