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Problem 1000. -- A+B Problem -- 衡阳八中OJ离线版-2014-11-04

1000: A+B Problem

Time Limit: 1 Sec  Memory Limit: 5 MB
Submit: 9981  Solved: 6311
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Description

Calculate a+b

Input

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

Output

Output a+b

Sample Input

1 2

Sample Output

3

HINT

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
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

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);
}
}

Source

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