PAT 甲级 1001  A+B Format

本文详细解析了PAT竞赛中经典题目A+B Format的解题思路与代码实现,介绍了如何使用字符数组处理数据,特别关注了零和负数情况的处理,提供了完整的Accepted代码示例。

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1001 A+B Format (20 point(s))

Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).

Input Specification:

Each input file contains one test case. Each case contains a pair of integers a and b where −10​6​​≤a,b≤10​6​​. The numbers are separated by a space.

Output Specification:

For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.

Sample Input:

-1000000 9

Sample Output:

-999,991

Experiential Summing-up 

this question....emmmm,I did cost a bit time to consider it . Mainly because I want to use string to cope with it . But it's complex to solve it by string type than using char array. So I did't find a better method until I refer to my last submit code. I find it so easy to format a ten size char array to process data. Certainly, we must handle two special cases which include when sum is equal to zero and when sum is negative number. There is nothing problem except what is said above. That's all~~╭(′▽`)╭(′▽`)╯

(The purpose of using English to portray my solution is that to exercise the ability of my expression of English and accommodate PAT advanced level's style.We can make progress together by reading and comprehending it. Please forgive my basic grammar's and word's error. Of course, I would appreciated it if you can point out my grammar's and word's error in comment section.( •̀∀•́ ) Furthermore, Big Lao please don't laugh at me because I just a English beginner settle for CET6    _(:з」∠)_  )

Accepted Code

#include <cstdio>

int main()
{
	int a,b;
	scanf("%d %d",&a,&b);
	int sum=a+b;
	char s[10];
	int len=0,x=0;
	if(sum==0)
		printf("0\n");
	else
	{
		if(sum<0)
			printf("-");
		sum=sum>0?sum:-sum;
		while(sum)
		{
			if(x!=0&&x%3==0)
				s[len++]=',';
			s[len++]=sum%10+'0';
			sum/=10;
			++x;
		}
		for(int i=len-1;i>=0;--i)
			printf("%c",s[i]);
		printf("\n");
	}
	return 0;
}

 

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