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

#include<bits/stdc++.h>
using namespace std;
const int N=1e3+10;
int a,b;
int main()
{
	cin >> a >> b;
	cout << a+b;
	return 0;
}

#include<bits/stdc++.h> using namespace std; const int N=1e3+10; int a,b; int main() { cin >> a >> b; cout << a+b; return 0; }

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

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

}

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

#include using namespace std; int main() { int a , b;

cin>>a>>b;

cout<<a+b;

return 0;	

}

#include<iostream>
using amespace std;
int vera(int x,int y){
	return x+y;
}
int main(){
	int a=1,b=2;
	cout<<vera(a,b);
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N=1e3+10;
int Accepted(int a,int b){//Accepted()
	return a+b; 
}
int main()
{
	int a,b;
	cin >> a >> b;
	cout << Accepted(a,b);
	return 0;
}
//老登布置的作业系列

#include<bits/stdc++.h>

using namespace std;
int main()
{
	string a1,b1;
	int a[10000]={},b[10000]={},c[10000]={};
	cin >> a1 >> b1;
	int lena=a1.size();
	int lenb=b1.size();
	for(int i=1;i<=lena;i++)
	{
		a[i]=a1[lena-i]-'0';
	} 
	for(int i=1;i<=lenb;i++)
	{
		b[i]=b1[lenb-i]-'0';
	}  
	int lenc=1;
	while(lenc<=lena||lenc<=lenb)
	{
		c[lenc]+=a[lenc]+b[lenc];
		if(c[lenc]>9)
		{
			c[lenc]-=10;
			c[lenc+1]++; 
		}
		lenc++;
	} 
	if(c[lenc]==0)
	{
		lenc--;
	}
	for(int i=lenc;i>=1;i--)
	{
		cout << c[i];
	}
	return 0;
}

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

A+B Problem题解

新用户强烈建议阅读此帖

首先我们要理清思路

1.需要用到什么样的头文件？

2.用什么样的数据范围？

3.思路是什么？

首先题目中的数据范围是1≤a,b≤10^6, 而int 的范围是-2147483648-2147483647 正合题意，所以数据类型可以用int

话不多说，直接上代码

#include<iostream>//导入头文件，iostream里面是标准输入输出流（我说的什么？） 
using namespace std;//使用标准命名空间 
int main(){//主函数，程序的入口 
	int a,b;//创建a,b两个整型变量 
	cin>>a>>b;//输入 a , b 两个变量 
	cout<<a+b; //输出a+b的内容 
	return 0; 
}

本蒟蒻发的第一篇题解，请多多支持喵~~

#include<bits/stdc++.h> using namespace std; const int N=1e3+10; int fx(int x,int y){ return x+y; } int main() { int a=1,b=2; cin>>a>>b; cout<<fx(a,b); return 0; }

#include<bits/stdc++.h> using namespace std; const int N=1e3+10; int fx(int x,int y){ return x+y; } int main() { int a=1,b=2; cin>>a>>b; cout<<fx(a,b); return 0; }

#include <stdio.h> #include using namespace std; int main() { int a,b; cin >> a >> b; cout << a+b << endl; }

谁不会这道题？？

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

其实是我不会

禁止发疯！！！

#include<bits/stdc++.h>//万能头文件
using namespace std;

int main(){
//定义int类型变量a,b
int a;
int b;
//输入变量a,b
scanf("%d",&a);
scanf("%d",&b);
//输出a,b
printf(" %d\n", a + b);
//exit(0); 或 return 0; 结束程序
return 0;
}

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

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

权威

#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;
}