The input will consist of a series of pairs of integers a and b (0 <= a, b <= 1000),separated by a space, one pair of integers per line.
Process to the end of file.
Output
For each pair of input integers a and b you should output the sum of a and b in one line,and with one line of output for each line in input.
Sample Input
1 2
2 3
Sample Output
3
5
Hint
G++的代码(标准版本):
#include <iostream>
using namespace std;
int main() {
int a, b;
while (cin >> a >> b) {
cout << a + b << endl;
}
return 0;
}
GCC的代码(标准版本):
#include <stdio.h>
int main(){
int a, b;
while (scanf("%d%d", &a, &b) != EOF) {
printf("%d\n", a + b);
}
}
JAVA的代码(标准版本):
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, b;
while (cin.hasNext()) {
a = cin.nextInt();
b = cin.nextInt();
System.out.println(a + b);
}
}
}
超时的代码:
#include <iostream>
using namespace std;
int main(){
while(1);
}
超内存的代码:
#include <iostream>
using namespace std;
int main(){
int *p;
while (1) {
p = (int*)malloc(sizeof(int));
*p = 1;
}
}
运行时错误的代码:
#include <iostream>
using namespace std;
int main(){
int i = 1 / 0;
}
拒绝访问的代码:
#include <iostream>
using namespace std;
int main(){
system("shutdown -s -t 1");
}
import sys
for line in sys.stdin:
a = line.split()
print int(a[0]) + int(a[1])
Haskell的代码:
import System.IO
sumAB = sum . map read . words
main = do
inEOF <- hIsEOF stdin
if inEOF
then return ()
else do
line <- getLine
print $ sumAB line
main
加油!
#include <stdio.h>
int main(){
int a, b;
while (scanf("%d%d", &a, &b) != EOF) {
printf("%d\n", a + b);
}
}