【ACM】ACM steps 1.1.3

A+B for Input-Output Practice (III)

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28263 Accepted Submission(s): 15357

Problem Description
Your task is to Calculate a + b.

Input
Input contains multiple test cases. Each test case contains a pair of integers a and b, one pair of integers per line. A test case containing 0 0 terminates the input and this test case is not to be processed.

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 5
10 20
0 0

Sample Output
6
30

解:

#include
 int main()
 {
    int a,b;
    scanf("%d %d",&a,&b);
    while(!(a==0&&b==0))
    {
    printf("%dn",a+b);
     scanf("%d %d",&a,&b);
    }
 }
华丽丽的一次过,没什么好说的……貌似之前的两题在wp上发表的时候格式有点问题……已经修正这个问题了~

《【ACM】ACM steps 1.1.3》上有2条评论

    1. (a==0&&b==0)就是a和b同时为0,这个条件不满足时前面的式子返回0,!(a==0&&b==0))就是!0,!0为真,执行循环体。而当前面的式子满足是返回1,!1为假,跳出循环体~

      Ps,炫优越+2

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