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

思路

我们可以把 $a$ 看作边 $(0,1)$ ， $b$ 看作 $(1,2)$ ，则只需要求 $(0,2)$ 的最短路即可。 考虑到边权可能为负，所以采用 floyd 算法。

code

#include <algorithm> //STL 通用算法
#include <bitset>//STL 位集容器
#include <cctype>         //字符处理
#include <cerrno>  //定义错误码
#include <cfloat> //浮点数处理
#include <ciso646>         //对应各种运算符的宏
#include <climits> //定义各种数据类型最值的常量
#include <clocale> //定义本地化函数
#include <cmath> //定义数学函数
#include <complex> //复数类
#include <csignal>         //信号机制支持
#include <csetjmp>         //异常处理支持
#include <cstdarg>         //不定参数列表支持
#include <cstddef>         //常用常量
#include <cstdio>  //定义输入／输出函数
#include <cstdlib> //定义杂项函数及内存分配函数
#include <cstring> //字符串处理
#include <ctime> //定义关于时间的函数
#include <cwchar>  //宽字符处理及输入／输出
#include <cwctype> //宽字符分类
#include <deque> //STL 双端队列容器
#include <exception> //异常处理类
#include <fstream>  //文件输入／输出
#include <functional>//STL 定义运算函数（代替运算符）
#include <limits>  //定义各种数据类型最值常量
#include <list>//STL 线性列表容器
#include <locale>//本地化特定信息
#include <map> //STL 映射容器
#include <memory>//STL通过分配器进行的内存分配
#include<new>//动态内存分配
#include <numeric>//STL常用的数字操作
#include <iomanip>  //参数化输入／输出
#include <ios> //基本输入／输出支持
#include <iosfwd>//输入／输出系统使用的前置声明
#include <iostream> //数据流输入／输出
#include <istream> //基本输入流
#include <iterator>        //STL迭代器
#include <ostream> //基本输出流
#include <queue> //STL 队列容器
#include <set> //STL 集合容器
#include <sstream> //基于字符串的流
#include <stack> //STL 堆栈容器
#include <stdexcept> //标准异常类
#include <streambuf> //底层输入／输出支持
#include <string>//字符串类
#include <typeinfo>        //运行期间类型信息
#include <utility> //STL 通用模板类
#include <valarray>        //对包含值的数组的操作
#include <vector>//STL 动态数组容器
using namespace std;
int edge[5][5];
int main(){
   memset(edge,0x1f,sizeof(edge));
   int a,b;
   cin>>a>>b;
   edge[0][1]=a;//建边
   edge[1][2]=b;//建边
   for(int k=0;k<=2;k++){//Floyd
      for(int i=0;i<=2;i++){
            for(int j=0;j<=2;j++){
               edge[i][j]=min(edge[i][j],edge[i][k]+edge[k][j]);
         }
      }
   }
   cout<<edge[0][2];
   return 0;
}

王者策划加强李信

#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;
}

很简单的一道题 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;
}

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

#include<bits/stdc++.h> using namespace std; int a[1000000]; int main() { int n,t; cin>>n; for(int i=1;i<=n;i++) { cin>>a[i]; } t=a[0]; for(int i=n;i>=1;i--)

	{
		
	cin>>a[i];
	cout<<a[i]<<" ";}
 	return 0; 
}

#include<bits/stdc++.h> using namespace std; int main() { int a,b,c,d,e,f,g,h; cin>>a>>b>>c; d=a/1000+a%1000/100+a%100/10+a%10; if(b>=10){ e=(b/10+b%10); } else{ e=b; } if(c>=10) { f=c/10+c%10; } else{ f=c; } g=d+e+f; h=g/10+g%10; cout<<h<<endl; return 0; }

各位大佬们不要在炸鱼为难新手了，这道题其实很简单 #include<iostream> using namespace std; int main() { int a,b; cin>>a>>b; cout<<a+b; return 0; }

#include<iosetream> using namespace std {int a,b int main() cin>>a>>b if(a>b) {cout<<"yes"<<endl } return0 }

转载自https://www.luogu.com.cn/problem/solution/P1001

~看来还没人用Splay，赶紧水一发~

因为加法满足交换律，所以我们可以把序列翻转一下，所求的总和不变

序列反转自然而然就能想到Splay啦

代码送上

//一颗资瓷区间加、区间翻转、区间求和的Splay
#include <bits/stdc++.h>
#define ll long long
#define N 100000
using namespace std;
int sz[N], rev[N], tag[N], sum[N], ch[N][2], fa[N], val[N];
int n, m, rt, x;
void push_up(int x){
    sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1;
    sum[x] = sum[ch[x][1]] + sum[ch[x][0]] + val[x];
}
void push_down(int x){
    if(rev[x]){
        swap(ch[x][0], ch[x][1]);
        if(ch[x][1]) rev[ch[x][1]] ^= 1;
        if(ch[x][0]) rev[ch[x][0]] ^= 1;
        rev[x] = 0;
    }
    if(tag[x]){
        if(ch[x][1]) tag[ch[x][1]] += tag[x], sum[ch[x][1]] += tag[x];
        if(ch[x][0]) tag[ch[x][0]] += tag[x], sum[ch[x][0]] += tag[x];
        tag[x] = 0;
    }
}
void rotate(int x, int &k){
    int y = fa[x], z = fa[fa[x]];
    int kind = ch[y][1] == x;
    if(y == k) k = x;
    else ch[z][ch[z][1]==y] = x;
    fa[x] = z; fa[y] = x; fa[ch[x][!kind]] = y;
    ch[y][kind] = ch[x][!kind]; ch[x][!kind] = y;
    push_up(y); push_up(x);
}
void splay(int x, int &k){
    while(x != k){
        int y = fa[x], z = fa[fa[x]];
        if(y != k) if(ch[y][1] == x ^ ch[z][1] == y) rotate(x, k);
        else rotate(y, k);
        rotate(x, k);
    }
}
int kth(int x, int k){
    push_down(x);
    int r = sz[ch[x][0]]+1;
    if(k == r) return x;
    if(k < r) return kth(ch[x][0], k);
    else return kth(ch[x][1], k-r);
}
void split(int l, int r){
    int x = kth(rt, l), y = kth(rt, r+2);
    splay(x, rt); splay(y, ch[rt][1]);
}
void rever(int l, int r){
    split(l, r);
    rev[ch[ch[rt][1]][0]] ^= 1;
}
void add(int l, int r, int v){
    split(l, r);
    tag[ch[ch[rt][1]][0]] += v;
    val[ch[ch[rt][1]][0]] += v;
    push_up(ch[ch[rt][1]][0]);
}
int build(int l, int r, int f){
    if(l > r) return 0;
    if(l == r){
        fa[l] = f;
        sz[l] = 1;
        return l;
    }
    int mid = l + r >> 1;
    ch[mid][0] = build(l, mid-1, mid);
    ch[mid][1] = build(mid+1, r, mid);
    fa[mid] = f;
    push_up(mid);
    return mid;
}
int asksum(int l, int r){
    split(l, r);
    return sum[ch[ch[rt][1]][0]];
}
int main(){
    //总共两个数
    n = 2;
    rt = build(1, n+2, 0);//建树
    for(int i = 1; i <= n; i++){
        scanf("%d", &x);
        add(i, i, x);//区间加
    }
    rever(1, n);//区间翻转
    printf("%d\n", asksum(1, n));//区间求和
    return 0;
}