题解 | 时间转换
时间转换
https://www.nowcoder.com/practice/c4ae7bcac7f9491b8be82ee516a94899
#include <iostream>
using namespace std;
int main() {
int seconds;
int a,b,c;
cin>>seconds;
if(seconds>3600){
a=seconds/3600;
b=(seconds-a*3600)/60;
c=(seconds-a*3600-b*60);
}
else {
a=0;
b=seconds/60;
c=seconds-b*60;
}
cout<<a<<" "<<b<<" "<<c<<endl;
return 0;
}
// 64 位输出请用 printf("%lld")
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