# How can i count the result?

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

1. ### marchanNew 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;
}```

Joined:
Jan 7, 2016
Messages:
2