#include<bits/stdc++.h> using namespace std; int main() { for(int i=0;i>=0;i++) { cout<<i<<" "; } return 0; }

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

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

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

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

鸡你太丑

#include <bits/stdc++.h>
 using namespace std; 
int main() { 
int a,b;
cin>>a>>b; 
cout<<a+b；//这题大家别想复杂了，不是高精度（我的噩梦😄 ）
return 0; 
}

嗨害嗨 来自菜鸽的第一篇题解。 由于本题是 萌新关爱 题 所以 本题的数据 不用高精度 就不多废话了 代码如下——

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

拜拜了友子们 我们下篇题解区见。

" #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; }

'c++'#include <bits/stdc++.h> 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(prelct)return true; return !(pre->son[1]==this||pre->san[0]this); } void node::pushdown() { if(thislct||!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); y->pre=x; 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("%c\n",query(B,A)); return 0; }

本题解转载自洛谷，并且没有经过授权。

原作者：Treeloveswater (Luogu User ID: 24559)

原题解链接

思路：使用了LCT(Link-Cut Tree)数据结构，高效运算A+B的值。

请不要复制粘贴

#include <bits/stdc++.h>
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->san[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);
y->pre=x;
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("%c\n",query(B,A));
return 0;
}