1:A+B Problem

总时间限制: 

1000ms 

内存限制: 

65536kB

描述

Calculate a + b

输入

Two integer a,,b (0 ≤ a,b ≤ 10)

输出

Output a + b

样例输入

1 2

样例输出

3

提示

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);
    }
}

源代码：

<pre name="code" class="cpp">int main(void)
{
    int a, b;
    scanf("%d%d", &a, &b);
    printf("%d", a+b);
    return 0;
}