BZOJ 官方题解：

Q: Where are the input and the output? A: Your program shall always read input from stdin (Standard Input) and write output to stdout (Standard Output). For example, you can use scanf in C or cin in C++ to read from stdin, and use printf in C or cout in C++ to write to stdout. You shall not output any extra data to standard output other than that required by the problem, otherwise you will get a Wrong Answer. User programs are not allowed to open and read from/write to files. You will get a Runtime Error or a Wrong Answer if you try to do so. Here is a sample solution for problem 1000 using C++/G++:

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

It's important that the return type of main() must be int when you use G++/GCC,or you may get compile error. Here is a sample solution for problem 1000 using C/GCC:

#include <stdio.h>
int main()
{
    int a,b;
    scanf("%d %d",&a, &b);
    printf("%d\n",a+b);
    return 0;
}

Here is a sample solution for problem 1000 using PASCAL:

program p1000(Input,Output);
var
  a,b:Integer;
begin
  Readln(a,b);
  Writeln(a+b);
end.

Here is a sample solution for problem 1000 using JAVA: Now java compiler is jdk 1.5, next is program for 1000

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);
    }
}

Old program for jdk 1.4

import java.io.*;
import java.util.*;
public class Main
{
    public static void main (String args[]) throws Exception
    {
        BufferedReader stdin =
        new BufferedReader(
        new InputStreamReader(System.in));
        String line = stdin.readLine();
        StringTokenizer st = new StringTokenizer(line);
        int a = Integer.parseInt(st.nextToken());
        int b = Integer.parseInt(st.nextToken());
        System.out.println(a+b);
    }
}

我来写一个Python题解。

a,b=map(int,input().split())#读入变量a、b，以空格隔开，并转为整型（int）
print(a+b)#输出a+b的值

不得不说Python是真心简单

C++

#include <cstdio>
#include <algorithm>
#include <iostream>

using namespace std;

int main(void)
{
    int a, b;
    scanf("%d%d", &a, &b);
    printf("%d\n", a + b);
    // cin >> a >> b;
    // cout << a + b;
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main(){
	int a,b;scanf("%d%d",&a,&b);
	printf("%d",a+b);
	return 0;
}

非常简单的一道题，输出a+b的和即可,跟H1001一样

代码：

#include<cstdio>//C语言风格输入输出头文件
int main(){ //主函数
    int a,b;//定义变量a和b
    scanf("%d%d",&a,&b);//输入a和b
    printf("%d",a+b);//输出a+b
    return 0; //返回值(一定要是0!)
}

水题一道。 这里大家可以看到世界各大语言的代码：

C

#include <stdio.h>

int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    printf("%d\n", a+b);
    return 0;
}

C++

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    int a,b;
    cin >> a >> b;
    cout << a+b << endl;
    return 0;
}

Pascal

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

Python

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

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);
    }
}

PHP

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

Ruby

a, b = gets.split.map(&:to_i)
print a+b

Rust

use std::io;

fn main(){
    let mut input=String::new();
    io::stdin().read_line(&mut input).unwrap();
    let mut s=input.trim().split(' ');

    let a:i32=s.next().unwrap()
               .parse().unwrap();
    let b:i32=s.next().unwrap()
               .parse().unwrap();
    println!("{}",a+b);
}

Go

package main

import "fmt"

func main() {
    var a, b int
    fmt.Scanf("%d%d", &a, &b)
    fmt.Println(a+b)
}

#include <bits/stdc++.h> using namespace std;
int main() {
int a, b;
cin >> a >> b;
cout << a + b;
return 0;
}
兄啊，这事给若至做的罢（恼

A + B Probelm 氵到要死

C++的入门题吧，输入 $a$和 $b$，输出$a+b$

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

c++版的：

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

太简单了 c++ AC code：

#include<bits/stdc++.h> //万能头
using namespace std; // 命名空间
int main() { //主函数main()
    int a,b; //定义整数a和b
    cin>>a>>b;//输入
    cout<<a+b;//输出
    return 0;//结束
}

很水的题。

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

题目中说了a、b都小于100，所以开int就行。

#include <iostream>

using namespace std;

int main()

{

long long A,B;//不开long long 见祖宗

cin>>A>>B;

cout<<A+B;

return 0;//不要我多讲了吧

}

球球了，点个👍吧，小哥哥帅气，小姐姐漂亮

C++11の高端的题解！

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<string>
#include<cstdlib>
using namespace std;//超长无关紧要的高端的头文件
int a,b;//定义高端的变量
int main(){
	scanf("%d%d",&a,&b);//快速读入高端的变量
	printf("%d",a+b);//经过简简单单的高端的计算后，快速输出高端的结果！
	return 0;//高端而潇洒的结尾~
}

看这高端的注释，不点个赞吗？

来一发dijkstra题解

#include<bits/stdc++.h>
using namespace std;
int n,m,s;
int head[100010],cnt=-1;
bool done[100010]={0};
long long ans[100010];
struct edge
{
    int to,next,w;
}e[200010];
void add(int u,int v,int w)
{
    e[++cnt].next=head[u];
    head[u]=cnt;
    e[cnt].to=v,e[cnt].w=w;
}
struct node
{
    long long num,value;
    bool operator <(const node &x) const{return value>x.value;}
    node(long long num_,long long value_):num(num_),value(value_){}
};
priority_queue<node>dis;
void dijkstra()
{
    while(!dis.empty())
    {
        node u=dis.top();
        dis.pop();
        if(done[u.num]==1)continue;
        done[u.num]=1;
        ans[u.num]=u.value;
        for(int i=head[u.num];~i;i=e[i].next)
        {
            int v=e[i].to;
            if(done[v])continue;
            dis.push(node(v,u.value+e[i].w));
        }
    }
}
int main()
{
    int a,b;
    cin>>a>>b;
    n=3,m=2,s=1;
    for(int i=1;i<=n;i++)
        head[i]=-1,dis.push(node(i,2147483647));
    add(1,2,a),add(2,3,b);
    dis.push(node(s,0));
    dijkstra();
    cout<<ans[3]<<endl;
    return 0;
}