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Question
Submitted by smurfs9109 on Sun, 2012-07-22 19:58
due on Thu, 2012-07-26 19:57
answered 1 time(s)
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C programming

C programming

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Answer
Submitted by Coloratus on Sun, 2012-07-29 19:03
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Complete solution

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xxxxxxxx<stdio.h>

xxxxxxx xxxxxxxxx xxx

xxx xxxxxxxxx xxxxxxxxxxxxxx

void xxxxxxxxxxxxx inputArray[], int xxx

xxx xxxxxxxxxxxxxxx xxxxxxxxxxxxx xxx n);

xxx xxxxxxxxxxxxxxxxx xxxxxxxxxxxxx xxx n);

void outputResults(int xxxxxxxx xxx maxN);

int xxxxxxxxxx

{

  int inputArray[MAX_INPUT + 1];    // Array in xxxxx xxxxx values xxx stored

xx xxx n;        xxxxxx xxxxxx     xxxxxx xxxxxx // xxxxxx of xxxxxxx xxxxxx before entering -1

xx xxx xxxxxxxxxxxxxx xxxxxx xxxxxx xxxxxx xxxxxx     xx Number xx xxxxxxxxx xxxxxx

  xxx maxN;                            xx xxx xxxxxxx xxxxxx xx marbles with same xxxxx

xx x x input(inputArray);

xx SortArray(inputArray, n);

xx nColors = xxxxxxxxxxxxxxxxxxxxxxx n);

  xxxx = xxxxxxxxxxxxxxxxxxxxxxxxx xxx

  outputResults(nColors, xxxxxx

xx

  xxxxxx xx

}

xxx input(int inputArray[])

x

  int xxxxxxxxxxxxxxxxx // xxxxxxxxx xxxxx variable

  int xxxxxxx x xxxxxxxx xx Counter xx entered values

xx do

xx x

xxxxxx xxxxxxxxxxxxx data xxx or -1 xx exit: xx xxxxxxx x xxx

      xxxxxxxxxxx &xxxxxxxxxxxx

      inputArray[counter++] = inputValue;

xx }while(inputValue xx xxxx

  xxxxxx xxxxxxxx x 1);

x

void SortArray(int xxxxxxxxxxxxx xxx n)

x

  int i;    // xxxxx xxxxxxxx

  xxx xxxxxxxx // xxxxxx xxxxxxxx

  int t;    // Temporary xxxxxxxx

xx xxxxx = xx x < x - xx xxxx

    xxxxx = x x 1; x < n; j++)

xxxxxxxxxx xxxxxxxxxxxxxxxx < inputArray[j])

      x

        t = inputArray[i];

        inputArray[i] x

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