#include<bits/stdc++.h>
using namespace std;
int main(){
int x,y;
cin>>x>>y;
if(0<=x,y<=32767){
cout<<x+y;
}else{
cout<<"NO";
}
return 0;
}
此题非常的简单
直接一个if()就可以过

特别简单

#include<bits/stdc++.h>//万能头文件
using namespace std;//使用一个标准的命名空间
int main(){
    int a,b;//定义变量a，b
    cin>>a>>b;//手动输入a和b
    cout<<a+b<<endl;//输出a+b的值
    return 0;//返回0 ，程序结束
}

点个赞再走呗🥺

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

很简单
#include<bits/stdc++.h>
using namespace std;
int main(){
int a,b;
cin>>a>>b;
cout<<a+b<<endl;
return 0;
}

#include <bits/stdc++.h>
using namespace std;
int main()
{
long long a,b,c;
cin>>a>>b;
c=a+b;
printf ("%d",c);
}

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

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

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

#include<bits/stdc++.h>
using namespace std;
signed main(){
    int a,b;
    cin>>a>>b;
    cout<<a+b;
}

来一篇好玩的题解。

#include <bits/stdc++.h>

#define _ using
#define __ namespace
#define ___ std;
#define ____ int
#define _____ main() {
#define ______ int a, b;
#define _______ cin >> a >> b;
#define ________ cout << a + b << '\n';
#define _________ }
_ __ ___
____ _____
    ______ _______ ________ _________

本小弟第一次发题解
#include <iostream>
using namespace std;
int main(){
long long a,b;//注意，不开long long见祖宗
cin>>a>>b;
cout<<a+b;//很水
}

暴力解题之进阶版

#include <iterator>
#include <functional>
#include<vector>
#include<deque>
#include<list>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<algorithm>
#include<numeric>
#include<memory>
#include<utility>
#define gou int main()
#define li {
#define guo int a,b;
#define jia cin>>a>>b;
#define sheng cout<<a+b;
#define si return 0;
#define yi }
using namespace std;
gou
li
guo
jia
sheng
si
yi


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

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstring>
using namespace std;
struct node
{
int data,rev,sum;
node *son[2],*pre;
bool judge();
bool isroot();
void pushdown();
void update();
void setson(node *child,int lr);
}lct[233];
int top,a,b;
node *getnew(int x)
{
node *now=lct+ ++top;
now->data=x;
now->pre=now->son[1]=now->son[0]=lct;
now->sum=0;
now->rev=0;
return now;
}
bool node::judge(){return pre->son[1]==this;}
bool node::isroot()
{
if(pre==lct)return true;
return !(pre->son[1]==this||pre->son[0]==this);
}
void node::pushdown()
{
if(this==lct||!rev)return;
swap(son[0],son[1]);
son[0]->rev^=1;
son[1]->rev^=1;
rev=0;
}
void node::update(){sum=son[1]->sum+son[0]->sum+data;}
void node::setson(node *child,int lr)
{
this->pushdown();
child->pre=this;
son[lr]=child;
this->update();
}
void rotate(node *now)
{
node *father=now->pre,*grandfa=father->pre;
if(!father->isroot()) grandfa->pushdown();
father->pushdown();now->pushdown();
int lr=now->judge();
father->setson(now->son[lr^1],lr);
if(father->isroot()) now->pre=grandfa;
else grandfa->setson(now,father->judge());
now->setson(father,lr^1);
father->update();now->update();
if(grandfa!=lct) grandfa->update();
}
void splay(node *now)
{
if(now->isroot())return;
for(;!now->isroot();rotate(now))
if(!now->pre->isroot())
now->judge()==now->pre->judge()?rotate(now->pre):rotate(now);
}
node *access(node *now)
{
node *last=lct;
for(;now!=lct;last=now,now=now->pre)
{
splay(now);
now->setson(last,1);
}
return last;
}
void changeroot(node *now)
{
access(now)->rev^=1;
splay(now);
}
void connect(node *x,node *y)
{
changeroot(x);
x->pre=y;
access(x);
}
void cut(node *x,node *y)
{
changeroot(x);
access(y);
splay(x);
x->pushdown();
x->son[1]=y->pre=lct;
x->update();
}
int query(node *x,node *y)
{
changeroot(x);
node *now=access(y);
return now->sum;
}
int main()
{
scanf("%d%d",&a,&b);
node *A=getnew(a);
node *B=getnew(b);
//连边 Link
connect(A,B);
//断边 Cut
cut(A,B);
//再连边orz Link again
connect(A,B);
printf("%d\n",query(A,B));
return 0;
}

#include<iostream>//头文件
using namespace std;//命名空间
int main()//主函数
{
    int a,b;//定义变量a b
        cin>>a>>b;//输入变量a b
        cout<<a+b<<endl;//输出a与b的和
    return 0;//返回值
}

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

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

最基础的A+B解法(给蒟蒻看的)

#include <iostream>//头文件很重要
using namespace std;
int main()//主函数
{
   int a,b,c;//定义三个变量a，b，c
   cin >> a >> b;//输入a，b值
   c = a+b;//求和并存入c中
   cout << c;//输出c
   return 0;
}

点个赞再走呗，跪谢！ヾ(≧▽≦*)o

#include <bits/stdc++.h>
using namespace std;

int main()
{
    long long a, b;
    
    scanf("%d %d", &a, &b);
    
    printf("%d", a + b);
    
    return 0;
}

#include <iostream>
#include <string>
using namespace std;
string BigNumAdd(string,int,string ,int);  //函数声明
int main()
{
    string a,b;       //用字符串来保存数据  C语言的朋友可以用char *
    cin>>a>>b;
    if(a.size()<b.size()) //作用:把长串放在a中  短串放在b中  最终结果是存在a中
    {
        string temp=a;
        a=b;
        b=temp;
    }
    cout<<BigNumAdd(a,a.size(),b,b.size())<<endl;  //函数调用
    return 0;
}
string BigNumAdd(string a,int lena,string b,int lenb)
{
    int aa,bb,sum,flag=0;       //flag进位标志，默认为0
    while(lena>0)
    {
        aa=a[lena-1]-'0';   //将a字符串的最后一个字符变成数字
        if(lenb>0)
            bb=b[lenb-1]-'0';  //将b字符串的最后一个字符变成数字
        else
            bb=0;
        sum=aa+bb+flag;     //sum用来保存a和b最后一个数字相加并加上进位
        if(sum>=10)         //相加大于10 当然要进位
        {
            a[lena-1]='0'+sum%10;
            flag=1;     //进位标志设为1
        }
        else
        {
            a[lena-1]='0'+sum;
            flag=0;
        }
        lena--;
        lenb--;
    }
    if(flag==1)   //如果最高位的前面还有进位
        a="1"+a;  //则字符串追加  把1追加到a字符串的前面
    return a;         //返回a作为 相加的结果
}