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1000: A+B Problem

Memory Limit：10 MB Time Limit：1.000 S
Judge Style：Text Compare Creator：
Submit：38 Solved：31

Description

Calculate a+b

Input

Two integer a,b (0<=a,b<=10)

Output

Output a+b

Sample Input Copy 

1 2

Sample Output Copy 

3

HINT

常见错误解答：
#include <iostream>
using namespace std ;     /* 上面两行不能改成 #include <iostream.h> */                   
void main()    /* 此处错误，main函数返回值必须是int，否则会编译错误 */
{
int a,b;
cout<<"Please input number a,b : "<<endl;  /*此处
错误，请一定注意题目要求，题目要求输出“两个整数的和”，所以只需输出一个整数即可，任何超出题目要求的输出（即使只是打印一行提示）都会导致 Wrong Answer ，因为系统是基于答案的完全匹配来评判的。*/
cin>>a>>b;
cout<<"a+b="<<a+b<<endl;  /*  "a+b="这一句是多余的，也是错误的，理由同上。*/
}

如果对程序答题格式不清楚的，可以查看系统左上角菜单的FAQ，里面有本题的各种语言的样例解答。在此列出用C/C++语言的答题示例如下（请把下面代码拷贝到代码提交框，点击提交）：

请注意代码中红颜色的部分，该部分与常用的VC编译器不同
C语言(使用GCC编译器）:
#include <stdio.h> 
int main() 
{ 
    int a,b; 
    scanf("%d %d",&a, &b); 
    printf("%d\n",a+b); 
    return 0; 
} 

C++语言（使用G++编译器）：
使用新的ANSI标准库，注意头文件的写法和using namespace  std 这句。
#include <iostream> 
using namespace std; 
int main() 
{ 
   int a,b; 
   cin >> a >> b; 
   cout << a+b << endl; 
   return 0; 
} 

另附：

注意:如果选择Java编译，类名要为Main，代码见下面。Java编译器为JDK 1.6。

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 to 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 to 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 to 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 to 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);
    }
}
Here is a sample solution to problem 1000 using Fortran: 
	PROGRAM P1000
		IMPLICIT NONE
		INTEGER :: A, B
		READ(*,*) A, B
		WRITE(*, "(I0)") A + B
	END PROGRAM P1000

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