回复 @ShallowDream雨梨 :
#include<cstdio>
#include<iostream>
using namespace std;
int main(){
int a,b=0;
freopen("reverse.in","r",stdin);
freopen("reverse.out","w",stdout);
cin>>a;
while (a!=0){
b=b*10+a%10;
a/=10;
}
cout <<b;
}
短不短？AC！！

一遍就AC了

回复 @2018noip必胜！ :
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
freopen("aplusb.in","r",stdin);
freopen("aplusb.out","w",stdout);
int a,b;
cin>>a>>b;
cout<<a+b<<endl;
return 0;
}
呵呵

渣渣程序#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main(void)
{ freopen("reverse.in","r",stdin);
freopen("reverse.out","w",stdout);
long long n = 0;int a[11]={0};
cin>>n;
if(n<0){
cout<<"-";n=-n;}
for(int i = 1;i++;)
{ if(a[i-1]==0&&n%10==0){a[i]=n%10;
n = n/10;continue;}
else{ a[i]=10;
cout<<n%10;n = n/10;if(n==0)break;
}
}
}我的该是最短了的吧

用int做的，没用字符串
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iomanip>
#include<cmath>
using namespace std;
int main(){
freopen("reverse.in","r",stdin);
freopen("reverse.out","w",stdout);
long long int n,n1,n2;
int a[12],t=0,t1=0;
cin>>n;
if(n<0) cout<<'-';
n=abs(n);
n2=n;
n1=n;
for(int i=1;;i++)
{n=n/10;
t++;
if(n==0) break;}
for(int i=1;i<=t;i++)
{a[i]=n1%10;
n1=n1/10;}
for(int i=1;;i++)
{if(n2%10==0) t1++;
n2=n2/10;
if(n2%10!=0) break;}
for(int i=1+t1;i<=t;i++)
cout<<a[i];
return 0;
}
很渣的程序 0.003s，仅供参考

勾起了我初学OI时的回忆..........

菜鸟提供的方案

guoshiju真的丑

放了3个月没做。。。我简直sb

没用字符串，他居然用负数坑我。