C++，我用个二进制吧。。。
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
int main()
{
int a,b,s=0,s1=0,i=0,na=0,nb=0;
cin>>a>>b;
if(a<=0) na=1,a*=-1;
while(a!=0)
{
if(a%2!=0)
s+=pow(2,a%2*i);
a/=2;
i++;
}
i=0;
if(na==1) s*=-1;
if(b<=0) nb=1,b*=-1;
while(b!=0)
{
if(b%2!=0)
s1+=pow(2,b%2*i);
b/=2;
i++;
}
if(nb==1) s1*=-1;
cout<<s+s1;;
return 0;
}

C答案，不谢，已AC
#include <stdio.h>
int main()
{
int x,y;
scanf("%d%d",&x,&y);
printf("%d",x+y);
return 0;
}

var a,b:longint;
begin
readln(a,b);
writeln(a+b);
end.

var n,k,i,num,sum,ans:longint; a:array[0..20] of longint;

function pangduan(n:longint):integer;
var xx,i:integer;
begin
xx:=trunc(sqrt(n+1));
for i:=2 to xx do
if (n mod i=0) then exit(0);
exit(1);
end;

procedure doit(t:longint);
var p,i:longint;
begin
if (num=k) then
begin
p:=pangduan(sum);
if (p=1) then inc(ans);
exit;
end;
for i:=t+1 to n do
begin
inc(num);
sum:=sum+a[i];
doit(i);
sum:=sum-a[i];
dec(num);
end;
end;

begin
readln(n,k);
for i:=1 to n do read(a[i]);
for i:=1 to n-k+1 do
begin
num:=1;
sum:=a[i];
doit(i);
end;
writeln(ans);
end.

冷清啊。。我来一发Python3的题解

a, b = map(int, input().split())
print(a + b)

恩 就是这么简洁qwq

#include <iostream>
using namespace std;
int main()
{
    int a,b;
    cin>>a>>b;
    cout<<a+b;
    return 0;
}

#include <iostream>
using namespace std;
int main()
{
int a,b;
cin>>a>>b;
cout<<a+b;
return 0;
}

#include<cstdio>
//调用头文件。
using namespace std;
int main()
{
int a,b;//定义a,b。
scanf("%d%d",&a,&b);//通过scanf格式化输入。
printf("%d",a+b);//直接输出表达式即可
return 0;
}

#include<iostream>
using namespace std;
int main()
{
int a,b,c;
cin>>a>>b;
c=a+b;
cout<<c<<endl;
return 0;
}

很简单

#include <cstdio>
#include <iostream>
using namespace std;
int main() {
    int a,b;
    cin>>a>>b;
    cout<<a+b<<endl;
    return 0;
}

#include<iostream>
#include<cstdio>
#include<ctime>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cstring>
using namespace std;
int main()
{
int A,B;
cin>>A>>B;
cout<<A+B;
return 0;
}

main函数自递归和位运算

#include<bits/stdc++.h>
int main(int a,int b,int k)
{
    if (k) scanf("%d%d",&a,&b);
    printf("%d",b==0?a:main(a^b,(a&b)<<1,0));
    exit(0);
}

#include <iostream>
using namespace std;
int main()
{
int a, b;
cin >> a >> b;
cout << a + b << endl;
return 0;
}

走一波面向对象的A+B。
<pre>
#include <bits/stdc++.h>

using namespace std;

class Something
{
friend istream & operator>>(istream & is, Something & as);
friend ostream & operator<<(ostream & os, Something & as);
public:
int a, b;
int sum(int x, int y);
Something & operator+(int z);
Something & operator+(Something b);

};

istream & operator>>(istream & is, Something & as)
{
is >> as.a;
return is;
}

ostream & operator<<(ostream & os, Something & as)
{
os << as.a;
return os;
}

int Something::sum(int x, int y)
{
return x + y;
}

Something & Something::operator+(int z)
{
a += z;
return *this;
}

Something & Something::operator+(Something b)
{
a += b.a;
return *this;
}

int main(int argc, char const *argv[])
{
Something sd, st;
cin >> sd >> st;
cout << sd + st;
return 0;
}
</pre>

#include<iostream> //基础头文件
using namespace std; //自定义函数
int main() //输入命令
{
int a,b=0,n=2; //1.定义a,用来输入两个数。 2.定义b,用来计数输入的两个数。 3.定义n,用来循环2次。
for(int i=1;i<=n;i++) //开始循环，循环次数n，n=2
{
cin>>a; //输入a；经循环共两次
b+=a; //计数，为了方便统计，输出时直接把b输出。
} //循环结束。
cout<<b; //输出循环后的结果。
return 0;
}

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

    #include <iostream>
    using namespace std;
    int main()
    {
    int a, b;
    cin >> a >> b;
    cout << a + b << endl;
    return 0;
    }

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

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
char a1[10000],b1[10000];int a[10000],b[10000],c[10000],x=0,lena,lenb,lenc=1,i;
int main()
{
    scanf("%s",a1);scanf("%s",b1);lena=strlen(a1);lenb=strlen(b1);
    for(i=0;i<=lena-1;i++) a[lena-i]=a1[i]-'0';
    for(i=0;i<=lenb-1;i++) b[lenb-i]=b1[i]-'0';
    while(lenc<=lena||lenc<=lenb)
    {
        c[lenc]=a[lenc]+b[lenc]+x;
        x=c[lenc]/10;
        c[lenc]%=10;
        lenc++;
    }
    if(0==(c[lenc]=x)) lenc--;
    for(i=lenc;i>=1;i--) cout<<c[i];return 0;
}

这道题实际上是一道最短路的模型题。我们只需要构造一个有三个顶点的无向图，1和2之间有一条边权为a的边，2和3之间有一条边权为b的边，而1和3之间有一条边权为maxlongint的边，那么答案就是1到3的最短路