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Problem 1000. -- A+B Problem

A+B Problem

Time Limit: 1 Sec  Memory Limit: 10 MB
Submit: 5188  Solved: 3471
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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

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

Source

POJ

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