1:A+B Problem
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- 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 1000import 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 1.4import 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;}
0 0
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