1001 A+B Format

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

. 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

源代码

//
// Created by sir Tian on 2019/1/10.
//
#include <stdio.h>

#include <math.h>

// 用于将数字转化为 字符数组
void getFormatInt(int num, char result[])
{

    int i;
    int remain = num;
    for (i = 0; remain > 0; i++)
    {
        result[i] = (remain % 10) + '0';
        remain /= 10;
    }
    result[i] = '\0';
}

int main()
{

    // factor_a,factor_b分别用来保存输入的两个相加的数，sum用于保存加数的和
    long factor_a, factor_b, sum = 0;

    // 两个数最大为10的六次方，其和最大不超过八位
    char ch[9];

    scanf("%ld %ld", &factor_a, &factor_b);

    // 两个数求和
    if (factor_a >= -pow(10, 6) && factor_a <= pow(10, 6) &&
        factor_b >= -pow(10, 6) && factor_b <= pow(10, 6))
    {
        sum = factor_a + factor_b;
    }

    // 如果和为负数则先把符号输出
    if (sum < 0)
    {
        printf("-");
        sum = 0 - sum;
    }

    // 将数的各个位保存到数组中
    getFormatInt(sum, ch);

    // 求数组的长度，便于控制在指定的位置输出分隔符号
    int i;
    for (i = 0; ch[i] != '\0'; i++)
    {
    }

    // 从数组后面向前输出
    for (int j = i - 1; j >= 0; j--)
    {
        printf("%c", ch[j]);
        // 从后往前数索引是三的倍数则需要输出一个分隔符，如果位置是0则不用输出
        if (j % 3 == 0 &&  j!=0)
        {
            printf(",");
        }
    }

    printf("\n");

    return 0;
}