如何从 C 中的数字中获取每个数字？

Question

I have these variables:

 int dividend;
 int divider;

And I have the function:

Divisibility7 (int num);

Those two variables will be at the main function, and the user will be asked to enter the dividend and divider, but in case the user enter the divider 7 , then, the function above will be called.

The problem is that I have to follow specific criteria to do this. So let's say the user will enter with the dividend 7203 . This happen:

I. Get the last digit of the number .

Last digit: 3

Ii. Multiply the last digit by 2

3 x 2 = 6

Iii. Get the value of the initial number , without the last digit.

720

Iv. Subtract the initial value without the last digit from the multiplication result.

fabs (720 - 6) = 714

V. Repeat the process until the result is a value less than or equal to 70

Vi. Compare the result with the table of contents (0, 7, 14, 21, 28, 35, 42, 49, 54, 63, 70) for Determine whether or not the number is divisible by 7

Code:

int res;
int x; 
int y;

int Divisibility7(int num) {

    int res;
int x;
int y;
int z;

    while(num > 70) {

    x = num % 10; // get the last digit from the number
    y = x * 2; // multiply the last digit by 2;
    z = num/10; // get the first digits from the number
    fabs(z - y); // subtract the first digits with the last digits;

    }

}

In the part of the while, the final fabs(zy) returns what I want, to be the first digits subtracting the last number, but the problem is that the while stop there, I have to do something to make this while go till 70 or less.

PS: I need to check if the final number from the iterations, it's a number multiplied by 7, how can I do that? And I can't use mod for this.

Answer 1

You have not changed num in your while loop . Also you do no return the value . Hope the following code will be ok for you .

int Divisibility7(int num) {

    int res,x,y,z;
    while(num > 70) {

    x = num % 10; // get the last digit from the number
    y = x * 2; // multiply the last digit by 2;
    z = num/10; // get the first digits from the number
    num = abs(z - y); // subtract the first digits with the last digits;

    }

 if(num == 0 || num == 7 || num == 14 || num == 21 || num == 28 || num == 35 || num == 42 || num == 49 || num == 54 || num == 63 || num == 70) {
    return 1;
}
else { 
    return 0;
}

}

Answer 2

Not sure, but I think this is what are you trying to do:

int main (void)
{
    int number, lastDigitMultiplied;
    scanf("%d", &number);
    while(number > 70){
        //get the last digit and multiply it by 2
        lastDigitMultiplied = (number % 10) * 2;
        //subtract the initial value without the last digit from the multiplication result.
        number = number / 10 - lastDigitMultiplied;
    }
    if(abs(number) % 7 == 0)
        printf("The result is %d and it is a multiple of 7", number);
    else
        printf("The result is %d and it is not a multiple of 7", number);
    return 0;
}

Answer 3

for divisibility against 7 , if you have a positive large bigint already formatted as a string , don't waste time with the regular method doing it 1 digit at a time.

powers of 10, mod 7 , repeats every 6 rounds:

10^1  %  7  =  3  10^7   %  7  =  3  10^13  %  7  =  3
10^2  %  7  =  2  10^8   %  7  =  2  10^14  %  7  =  2
10^3  %  7  =  6  10^9   %  7  =  6  10^15  %  7  =  6

10^4  %  7  =  4  10^10  %  7  =  4  10^16  %  7  =  4
10^5  %  7  =  5  10^11  %  7  =  5  10^17  %  7  =  5
10^6  %  7  =  1  10^12  %  7  =  1  10^18  %  7  =  1

meaning, the effects of the digits mod 7 stay identical as long as they're in chunks of 6 digits at a time.

the largest multiple of 6-digits supported by double precision FP to full integer precision would be 12-digits at a time.

So the way to do it is to right-align the digits by padding leading edge zeros to ensure string length is a multiple of 12 . Then simply add the digits onto each other, performing a single mod % 7 operation once every 9000 or so rounds of the addition loop

—- (that's when you run up against 2^53 limit; 9007 rounds if u wanna be pedantic about it)

example:

   x = 71400535477047763120175175402859619447790
       02233464423375355339031113233806609150957

x % 7 = 4

now try summing up chunks of 12:

007140053547704776312017517540285961944779
002233464423375355339031113233806609150957

007140053547      7140053547
704776312017    711916365564
517540285961   1229456651525
944779002233   2174235653758
464423375355   2638659029113
339031113233   2977690142346
806609150957   3784299293303
----------------------------
               3784299293303 % 7 = 4

it works exactly the same for any multiples of 6: eg 6, 18, 24, and 30 --

00714005354770477631201751754
02859619447790022334644233753
55339031113233806609150957

312017   312017
517540   829557
285961  1115518
944779  2060297
002233  2062530
464423  2526953
375355  2902308
339031  3241339
113233  3354572
806609  4161181
150957  4312138
007140  4319278
053547  4372825
704776  5077601
-----   -------
_       5077601 % 7 = 4

00000000714005354770477631201
75175402859619447790022334644
23375355339031113233806609150957

464423375355339031   464423375355339031
113233806609150957   577657181964489988
000000007140053547   577657189104543535
704776312017517540  1282433501122061075
285961944779002233  1568395445901063308
--------------------------------------- 
                    1568395445901063308 % 7 = 4

00000000000000714005354770477
63120175175402859619447790022
33464423375355339031113233806
609150957

339031113233806609150957   339031113233806609150957
000000000000007140053547   339031113233813749204504
704776312017517540285961  1043807425251331289490465
944779002233464423375355  1988586427484795712865820
---------------------------------------------------
                          1988586427484795712865820  % 7 = 4

000000007140053547704776312017517
540285961944779002233464423375355
339031113233806609150957

000000007140053547704776312017          7140053547704776312017
517540285961944779002233464423  517540293101998326707009776440
375355339031113233806609150957  892895632133111560513618927397

                                892895632133111560513618927397  % 7 = 4