1001. A+B Format (20) ——PAT (Advanced Level) Practise

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本文介绍了一个编程问题，即计算两个整数a和b的和，并将结果以标准格式输出，即每三位数字用逗号分隔。文章提供了一个C语言程序示例，该程序能够处理包括负数在内的各种输入情况。

题目信息：

1001. A+B Format (20)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.

Output

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

代码如下：

#include <stdio.h>
int main()
{
	int a, b;
	char arr[30], flag;
	arr[29] = '\0';
	while (scanf("%d%d", &a, &b) != EOF)
	{
		int k = 29;
		int s = a + b, t = 0;
		flag = 0;
		if (s < 0)
		{
			flag = 1;
			s *= -1;
		}
		if (s == 0)
			arr[--k] = '0';
		while (s)
		{
			arr[--k] =  '0' + s % 10;
			if (++t == 3)
			{
				t = 0;
				if (s >= 10)
					arr[--k] = ',';
			}
			s /= 10;
		}
		if (flag)
			arr[--k] = '-';
		printf("%s\n", arr + k);
	}
	return 0;
}