将一个9位数转换为三个3位数

Question

Ok everyone so i have a 9 digit number in input ( 123456789 ) and i need to break it down to first middle and last three numbers ( 123,456,789 ). I know that i can use modulo to get the last three digits.

first_group_of_three = input number % 1000;

But i have no idea how to get the other two groups. Any help would be appreciated.

UPDATE ! Thanks to milevyo answer i managed to figure it out.Now for the code freaks out there,i am a beginner programmer and i would like to ask what is a more efficient way to write this code(below),maybe using less variables or some other technique.I didn't transform it to char because we are not allowed to since we still "havent learned it"

     scanf("%d" , &input_number);
   last_group_of_three = input_number % 1000;
   temp = (input_number-last_group_of_three)/1000;
   middle_group_of_three = temp % 1000;
   first_group_of_three = (temp-middle_group_of_three)/1000;

Answer 1

first_group_of_three = input number % 1000;
shift=(input number-first_group_of_three)/1000

second_group_of_three = shift % 1000;

and so on

Answer 2

Simplification: @Barmar

scanf("%d", &input_number);

last_group_of_three = input_number % 1000;
input_number /= 1000;

middle_group_of_three = input_number % 1000;
input_number /= 1000;

// % 1000 only if original input exceeded 9 digits
first_group_of_three = input_number % 1000; 

Note: many compilers will optimize the % 1000 and / 1000 into a single operation providing the two results: remainder & quotient.

(OP has not mention what code should do with negative numbers.)

Answer 3

您可以在char表中以字符串形式输入数字，然后前3个数字将保留在表的第4个char中，成为char'，'，第5、6、7th个char将是下3个数字，以此类推。最后创建字符串“ 123,456,789”，然后打印。