C :

#include "stdio.h"

int main(int argc, char* argv[])
{
	int n,a,b,c;
	scanf ("%d",&n);
	a=(n/3600)%24;
	b=(n-n/3600*3600)/60%60;
	c=(n-n/3600*3600-(n-n/3600*3600)/60*60)%60;
	printf("%02d:%02d:%02d\n",a,b,c);
}

C++ :

#include <iostream>  
using namespace std;  
int main(){  
   int n;
   cin>>n;
   int a = n / 3600;
   int b = n % 3600 / 60;
   int c = n % 3600 % 60;
   if(a < 10){ 
      cout<<"0"<<a<<":";
   }else{ 
      cout<<a<<":";
   }
   if(b < 10){ 
      cout<<"0"<<b<<":";
   }else{ 
      cout<<b<<":";
   }
   if(c < 10){ 
      cout<<"0"<<c<<endl;
   }else{ 
      cout<<c<<endl;
   }
   return 0;  
} 

Python :

n = int(input())
shi = 0
fen = 0
miao = 0
t = 0
# 如果n整除3600大于等于1,则取整赋值给shi,余数赋值给t,t再分两种情况分析,一种是大于等于60,另外就是小于60
if n // 3600 >= 1:
    shi = n // 3600
    t = n % 3600
    #余数大于60的情况
    if t // 60 >= 1:
        fen = t // 60
        miao = t % 60
    #余数小于60的情况
    else:
        miao = t
#n小于3600时
else:
    fen = n // 60
    miao = n % 60
#格式化输出
print('%02d:%02d:%02d' % (shi, fen, miao))