#include<iostream>
using namespace std;
int main(){
long long a,b,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z;
cin>>a>>b;
long long c=a+b;
cout<<c;
return 0;
}

这题是入门的基础题，可以用很多方法求解，下面是最简单的几种方法，讲解与代码奉上。

1.普通解法：运用了普通计算机加法计算方法，C与C++代码如下：

//c语言解法
#include<cstdio>
using namespace std;
int main(){
    int a, b;
    scanf("%d%d", &a, &b);
    printf("%d%d", a + b);
    return 0;
}

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

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2.LCT(Link-Cut Tree)解法：使用动态树来解这题，是较~简单~难的，不多说，代码奉上：：

#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);
        connect(A,B);
        cut(A,B);
        connect(A,B);
    printf("%d\n",query(A,B)); 
    return 0;
}

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3.树状数组解法：也很“简单”，代码如下：

#include<iostream>
#include<cstring>
using namespace std;
int lowbit(int a)
{
    return a&(-a);
}
int main()
{
    int n=2,m=1;
    int ans[m+1];
    int a[n+1],c[n+1],s[n+1];
    int o=0;
    memset(c,0,sizeof(c));
    s[0]=0;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        s[i]=s[i-1]+a[i];
        c[i]=s[i]-s[i-lowbit(i)];
    }
    for(int i=1;i<=m;i++)
    {
        int q=2;
        if(q==1)
        {
            int x,y;
            cin>>x>>y;
            int j=x;
            while(j<=n)
            {
                c[j]+=y;
                j+=lowbit(j);
            }
        }
        else
        {
            int x=1,y=2;//求a[1]+a[2]的和
            int s1=0,s2=0,p=x-1;
            while(p>0)
            {
                s1+=c[p];
                p-=lowbit(p);
            }
            p=y;
            while(p>0)
            {
                s2+=c[p];
                p-=lowbit(p);
            }    
            o++;
            ans[o]=s2-s1;
        }
    }
    for(int i=1;i<=o;i++)
        cout<<ans[i]<<endl;
    return 0;
}

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4.Splay解法：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，代码如下：

#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(){
    n = 2;
    rt = build(1, n+2, 0);
    for(int i = 1; i <= n; i++){
        scanf("%d", &x);
        add(i, i, x);
    }
    rever(1, n);
    printf("%d\n", asksum(1, n));
    return 0;
}

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5.Dijkstra+STL的优先队列优化：代码如下：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <climits>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
#include <ctime>
#include <string>
#include <cstring>
using namespace std;
const int N=405;
struct Edge {
    int v,w;
};
vector<Edge> edge[N*N];
int n;
int dis[N*N];
bool vis[N*N];
struct cmp {
    bool operator()(int a,int b) {
        return dis[a]>dis[b];
    }
};
int Dijkstra(int start,int end)
{
    priority_queue<int,vector<int>,cmp> dijQue;
    memset(dis,-1,sizeof(dis));
    memset(vis,0,sizeof(vis));
    dijQue.push(start);
    dis[start]=0;
    while(!dijQue.empty()) {
        int u=dijQue.top();
        dijQue.pop();
        vis[u]=0;
        if(u==end)
            break;
        for(int i=0; i<edge[u].size(); i++) {
            int v=edge[u][i].v;
            if(dis[v]==-1 || dis[v]>dis[u]+edge[u][i].w) {
                dis[v]=dis[u]+edge[u][i].w;
                if(!vis[v]) {
                    vis[v]=true;
                    dijQue.push(v);
                }
            }
        }
    }
    return dis[end];
}
int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    Edge Qpush;

    Qpush.v=1;
    Qpush.w=a;
    edge[0].push_back(Qpush);

    Qpush.v=2;
    Qpush.w=b;
    edge[1].push_back(Qpush);

    printf("%d",Dijkstra(0,2));
    return 0;
}

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6.模拟：模拟人工运算方法，代码如下：

#include <iostream> 
#include <cmath>
using namespace std;
int fu=1,f=1,a,b,c=0;
int main()
{
    cin>>a>>b;
    if(a<0&&b>0)fu=2;
    if(a>0&&b<0)fu=3;
    if(a<0&&b<0)f=-1;
    if(a==0){cout<<b;return 0;}
    if(b==0){cout<<a;return 0;} 
    a=abs(a);
    b=abs(b);
    if(a>b&&fu==3)f=1;
    if(b>a&&fu==3)f=-1;
    if(b>a&&fu==2)f=1;
    if(b<a&&fu==2)f=-1;
    if(fu==1)c=a+b;
    if(fu>1)c=max(a,b)-min(a,b);
    c*=f;
    cout<<c;
    return 0;
}

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7.字典树：代码如下：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
struct node{
    int str[26];
    int sum;
}s[1000];
char str1[100];
int t=0,tot=0,ss=0;
bool f1;
void built()
{
    t=0;
    for(int i=0;i<strlen(str1);i++)
    {
         if(str1[i]=='-'){
             f1=true;continue;
         }
         if(!s[t].str[str1[i]-'0'])
         s[t].str[str1[i]-'0']=++tot;
         t=s[t].str[str1[i]-'0'];
         s[t].sum=str1[i]-'0';
    }
}
int query()
{
   int t=0;int s1=0;
   for(int i=0;i<strlen(str1);i++)
   {
           if(str1[i]=='-') continue;
           if(!s[t].str[str1[i]-'0']) return s1;
           t=s[t].str[str1[i]-'0'];
           s1=s1*10+s[t].sum;
   }
   return s1;
}
int main()
{    
  for(int i=1;i<=2;i++)
  {
      f1=false;
      scanf("%s",str1);
    built();
    if(f1)
      ss-=query();
      else ss+=query();
  }
  printf("%d",ss);
  return 0;    
}

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8.二进制：用二进制的计算方法计算，代码如下：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
int main()
{
    int a,b,s=0,s1=0,i=0,na=0,nb=0;
    cin>>a>>b;
    if(a<=0) na=1,a*=-1;
    while(a!=0)
    {
        if(a%2!=0)
        s+=pow(2,a%2*i);
        a/=2;
        i++;
    }
    i=0;
    if(na==1) s*=-1;
    if(b<=0) nb=1,b*=-1;
    while(b!=0)
    {
        if(b%2!=0)
        s1+=pow(2,b%2*i);
        b/=2;
        i++;
    }
    if(nb==1) s1*=-1;
    cout<<s+s1;;
    return 0;
}

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9.最小生成树：代码如下：

#include <cstdio>
#include <algorithm>
#define INF 2140000000
using namespace std;
struct tree{int x,y,t;}a[10];
bool cmp(const tree&a,const tree&b){return a.t<b.t;}
int f[11],i,j,k,n,m,x,y,t,ans;
int root(int x){if (f[x]==x) return x;f[x]=root(f[x]);return f[x];}
int main(){
    for (i=1;i<=10;i++) f[i]=i;
    for (i=1;i<=2;i++){
        scanf("%d",&a[i].t);
        a[i].x=i+1;a[i].y=1;k++;
    }
    a[++k].x=1;a[k].y=3,a[k].t=INF;
    sort(a+1,a+1+k,cmp);
    for (i=1;i<=k;i++){
        x=root(a[i].x);y=root(a[i].y);
        if (x!=y) f[x]=y,ans+=a[i].t; 
    }
    printf("%d\n",ans);
}

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其他语言解法

1.Pascal

var a, b: longint;
begin
    readln(a,b);
    writeln(a+b);
end.

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2.Python2

s = raw_input().split()
print int(s[0]) + int(s[1])

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3.Python3

s = input().split()
print(int(s[0]) + int(s[1]))

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4.Java

import java.io.*;
import java.util.*;
public class Main {
    public static void main(String args[]) throws Exception {
        Scanner cin=new Scanner(System.in);
        int a = cin.nextInt(), b = cin.nextInt();
        System.out.println(a+b);
    }
}

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5.JavaScript （Node.js）

const fs = require('fs')
const data = fs.readFileSync('/dev/stdin')
const result = data.toString('ascii').trim().split(' ').map(x => parseInt(x)).reduce((a, b) => a + b, 0)
console.log(result)
process.exit()

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6.php

<?php
$input = trim(file_get_contents("php://stdin"));
list($a, $b) = explode(' ', $input);
echo $a + $b;

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---

以上是我想出来的亿种方法，希望大家点个赞支持一下

不需要提示

直接 scanf printf，不需要输入和输出的提示

请不要输出“输入”和“输出”

使用$\colorbox{yellow}{Markdwon}$以获得更加效果