PAT甲级1001. A+B Format

约定不等于承诺〃 2022-06-11 02:26 144阅读 0赞

*  A+B Format (20)

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

### 题目要求： ###

计算a+b然后格式化输出结果，格式化输出要求：结果如果超过四位，就用小数点每三位隔开输出；

输入：输入两个在\[-1000000,1000000\]的a,b两个数

输出：应该在一行格式化输出

### 解题思路： ###

a,b的和的取值范围为\[-1000000,1000000\]没有超过C语言int类型数据的范围，所以不用担心溢出的问题，结果每三位需要输出一个逗号，将结果不断除1000取余，就会没一次得到少于三位的余数，需要对余数进行判断，不足三位的需要用0补足最高位，但最后一次取余不需要用零来补位。

#include "stdio.h"
    #include "stdlib.h"
    
    int main() {
    
        int a = 0;
        int b = 0;
        int c = 0;
        int num = 0;
        int tmp[3];
    
        scanf("%d%d", &a, &b);
        c = a + b;
    
        if (c < 0)
        {
            printf("%c", '-');
            c = -c;
        }
    
        while (c/1000>=1)
        {
            tmp[num] = c % 1000;
            num++;
            c = c / 1000;
    
        }
        tmp[num] = c;
    
        for (int  i = num; i >=0; i--)
        {
            if (i==num)
            {
                printf("%d", tmp[i]);
            }
            else
            {
                if (tmp[i]<10)
                {
                    printf("%s%d", ",00", tmp[i]);
                }
                else if (tmp[i]<100)
                {
                    printf("%s%d", ",0", tmp[i]);
                }
                else
                {
                    printf("%s%d", ",", tmp[i]);
                }
    
            }
        }
        return 0;
    }