思路

根据 $3600 \times h + 60 \times m +s$ 推即可。 不开 long long 见祖宗

code

#include<bits/stdc++.h>
using namespace std;
int main(){
   long long h,m,s;
   cin>>h>>m>>s;
   cout<<1ll*h*3600+m*60+s;//1ll表示转化 long long
   return 0;
}

#include<bits/stdc++.h> 
using namespace std;int main(){long long int a,b,c; cin>>a>>b>>c; cout<<a*3600+b*60+c; return 0; }

两行代码就够了

#include<bits/stdc++.h> using namespace std; int main() { int h,m,s,t; cin>>h>>m>>s; t=3600h+60m+s; cout<<t; return 0; }

#include<bits/stdc++.h> using namespace std; int main() { int h,m,s,t; cin>>s; h=3600s; m=60s; t=h+m+s; cout<<t<<endl; return 0; }

#include<bits/stdc++.h> using namespace std; int main() { int a,b,c,d,e,f; cin>>a>>b>>c; d=a6060; e=b*60; f=c+d+e; cout<<fixed<<setprecision(0)<<f; return 0; }

#include<iostream> using namespace std; int main() { int h,m,s,a,b,c,d; cin>>h>>m>>s; a=h3600; b=m60; c=s; d=a+b+c; cout<<d; return 0; }