我直接复制了高精度的代码

#include <bits/stdc++.h>
using namespace std;
string a,b;
int x[100000],y[100000];
int z[100000];
int main(){
	cin>>a>>b;
	int len=max(a.size(),b.size());
	//逆置转整 
	for(int i=a.size()-1,j=0;i>=0;i--,j++){
		x[j]=a[i]-'0';
	}
	for(int i=b.size()-1,j=0;i>=0;i--,j++){
		y[j]=b[i]-'0';
	}
	//加
	for(int i=0;i<len;i++){
		z[i]+=x[i]+y[i];
		if(z[i]>=10){
			z[i]-=10;
			z[i+1]+=1;
		}
	}
	//输出
	int f=0;//无效输出 
	for(int i=len;i>=0;i--){
		if(z[i]!=0){
			f=1;
		}
		if(f==1){
			cout<<z[i];
		}
	}
	return 0; 
}

~~#include #include

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<<endl; return 0; }

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

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

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

代码高亮，简洁而有效的注解，思路清晰，便于理解，无冗长板子，不定义非常用宏。

说的太好了

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

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

1

A+B Problem 题解

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

推荐题目

本蒟蒻啥都不会，只能贡献一份小小的码风优良的代码```language

#include <iostream>
using namespace std;

const int MAX = 1000;

void change(const string &a1, int a[], int len)
{
	for (int i = 0; i < len; i++)
	{
		a[i] = a1[len - i - 1] - '0';
	}
}

int main()
{
	string a1, b1;
	cin >> a1 >> b1;
	int lena = a1.size(), lenb = b1.size();
	int a[MAX] = {}, b[MAX] = {}, c[MAX] = {}, cf = 0;
	change(a1, a, lena);
	change(b1, b, lenb);
	int lenc = lena;
	if (lenb > lena)
	{
		lenc = lenb;
	}
	for (int i = 0; i <= lenc; i++)
	{
		c[i] = a[i] + b[i] + cf;
		if (c[i] >= 10)
		{
			cf = 1;
			c[i] -= 10;
		}
		else
		{
			cf = 0;
		}
	}
	int i = lenc;
	if (c[i] == 0)
	{
		i--;
	}
	for (i; i >= 0; i--)
	{
		cout << c[i];
	}
	return 0;
}

#include<bits/stdc++.h> //头文件
using namespace std;//命名空间
int main(){//主函数,程序从这里开始
    //signed main也行
    long long a,b;
    //用long long类型的变量进行存储因为数值范围是10^6，int存不下
    cin>>a>>b;
    /*
      这里可以替换为：
      scanf("%lld%lld",&a,&b);
    */
    long long ans=a+b;
    cout<<ans<<endl;
    /*
      这里可以替换为：
      printf("%lld",ans);
    */
    return 0;//一定要写,不然RE
}

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

终于可以写一份A+B这么难的题的题解了。
咦？竟然没有人写LCT的题解？
Link-Cut Tree 会很伤心的！
ORZ为了不让LCT伤心于是我来一份LCT的A+B题解吧！
送上代码：

#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;
}

本蒟蒻第一次写题解，请原谅

#include<iostream>//头文件
using namespace std;//使用标准命名空间
int main()//主函数{
	int a,b;
	cin>>a>>b;
	cout<<a+b;
	return 0;//好习惯，拜拜
}

这是全站最简单的题，也是所有萌新都会做到的题。

#include<bits/stdc++.h>
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;
}

line = list(map(int,input().split()))
print(line[0]+line[1])

#include<bits/stdc++.h>
using namespace std;
const int N=100001;
int a[N],b[N],ml;
void add(int a[],int b[])
{
	int cnt=0;
	for(int i=1;i<=ml+1;i++)
	{
		cnt+=a[i]+b[i];
		a[i]=cnt%10;
		cnt/=10;
	}
	return;
}
void print(int a[])
{
	int i;
	for(i=ml+1;i>0&&!a[i];i--);
	if(!i)
	{
		cout<<0;
		return;
	}
	while(i)
	{
		cout<<a[i--];
	}
}
int main()
{
	string sa,sb;
	cin>>sa>>sb;
	int la=sa.size(),lb=sb.size();
	ml=la>lb?la:lb;
	for(int i=1;i<=la;i++)
	{
		a[i]=sa[la-i]-'0';
	}
	for(int i=1;i<=lb;i++)
	{
		b[i]=sb[lb-i]-'0';
	}
	add(a,b);
	print(a);
	return 0;
}