PAT Grade A A1001 A+B Format (20 points) (Question + code + detailed comment + two methods to achieve)

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

//法一:按照要求每三位一组的输出，之要注意输出，的时机即可
#include<cstdio>
int main() {
    int a, b;
    scanf("%d%d", &a, &b);
    int sum = a + b;
    if (sum < 0) {
        printf("-");        //如果是负数，先输出负号，再转为正数
        sum *= -1;
    }
    int s[10];          //数组s储存和的每一位数字
    int len = 0;
    do {
        s[len++] = sum % 10;
        sum /= 10;
    } while (sum);        //用do-while而不用while是为了防止有sum = 0的情况

    for (int i = len - 1; i >= 0; i--) {         //因为先得到的是原数的低位，所以要逆序输出
        printf("%d", s[i]);
        if (i && i % 3 == 0)         //第一个条件是因为当数字位数刚好是3的倍数时，最后三位数后面不用输出，
            printf(",");
    }
    
    return 0;
}

//Method 2: Make good use of the range of a and b in the title , sum = a + b, then the range of sum is (2e-6)~(2e+6), then the output can be based on the size of sum, nothing more than three Two situations : (1) sum >= 1e+6 , that is, the number of sum digits is greater than 6 and less than 7, then the last 6 digits are output in two groups (that is, two commas), and then the extra 1 digit in front is output; (2) 1000 < =sum <= 1e+6 , there is only one comma, and the others will directly output (3) sum <1000 , then the digits of sum are less than or equal to 3, just output sum directly without commas. I believe that the meaning is easy to understand , but the writing style of this 蒻蒻 is limited, and the expression may not be so clear. Please forgive me! ! Here is the code of AC :

#include<stdio.h>
int main() {
    int a, b;
    scanf("%d%d", &a, &b);
    int ans = a + b;
    if (ans < 0) {         //对负数处理和第一种方法一样
        printf("-");
        ans *= -1;
    }
    if (ans >= 1000000)     //三种可能的情况
        printf("%d,%03d,%03d", ans / 1000000, ans % 1000000 / 1000, ans % 1000);
    else if (ans >= 1000)
        printf("%d,%03d", ans / 1000, ans % 1000);
    else
        printf("%d", ans);
    return 0;
}