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How can i count the result?

Discussion in 'C' started by marchan, Jan 7, 2016.

  1. marchan

    marchan New Member

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    hi guys, this the code i found in the net but i did a little changes, my question is how can i count the result at the IF part?
    Code:
    #include <stdio.h>
    #include <math.h>
    int combinationUtil(int arr[], int data[], int start, int end, int index, int r,float MEAN);
    int printCombination(int arr[], int n, int r,float MEAN)
    {
       
        int data[r];
    combinationUtil(arr, data, 0, n-1, 0, r,MEAN);
    }
    int combinationUtil(int arr[], int data[], int start, int end, int index, int r,float MEAN)
    {
        int j,i,a=1,b=0,f=0,ch;
        float sum=0,mean,deviation=0,standard_deviation;
    
    
       
        if (index == r){
    
    
            for (j=0; j<r; j++)
            {
               sum=sum+data[j];
                deviation=deviation+pow((MEAN-data[j]),2);
            }
            printf("\n");
           mean=sum/r;
           standard_deviation=sqrt(deviation/(r-1));
     
            if(MEAN<=(mean+(2.776*standard_deviation))&& MEAN>=(mean-(2.776*standard_deviation)))//{
            printf("%d",a);
         
            else printf("%15d",b); 
      
           return;}
    //}
    
     //fprintf(fin,"\n");
    
           // int c= f;
    
    
       
       for (i=start; i<=end && end-i+1 >= r-index; i++)
        {
    
            data[index] = arr[i];
            combinationUtil(arr, data, i+1, end, index+1, r,MEAN);
    
        }
     
    
    
    }
    
    
    int main()
    {
        int arr[] = {2,4,6,8,10,12,16};
        int r = 4,i,countt=0;
        int n = sizeof(arr)/sizeof(arr[0]);
        float sum=0,MEAN;
        for(i=0; i<n; i++)
        {
            sum=sum+arr[i];
        }
        MEAN=sum/n;
        printf("MEAN=%f\n\n\n",MEAN);
        printCombination(arr, n, r,MEAN);
        return 0;
    }
     
  2. marchan

    marchan New Member

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    is there anybody who wants to help me?
     

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