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  # array
  
  given an array x[10][10] with initial values, compute the sum of all:-
  b)the square of the elements in the even rows(i.e., rows 2,4,6,8,10)
  c)the square root of the elements in the odd columns(i.e.,columns 1,3,5,7,9).
  0
  samuelccs
  1 Solution
  
  Commented:
  #include <math.h>
  
  //assuming x[row][column]
  //s is squares, sr is square root, return value is diagonal elements
  
  int calculate(int *x[10][10], int *s, int *sr) {
  int i,j,k, sum=0;
  *s=0; *sr=0;
  for(i=0,j=9;i<10;i++ ,j--){
  sum+=*x[i][i]+*x[i][j];
  for(k=0;k<9;k++) if(i%2)      *s+=*x[i][k]**x[i][k]; else *sr+=sqrt(*x[k][i]);
  }
  return(sum-=*x[4][4]);
  }
  0
  
  Commented:
  Hey ScrapDog, keep doing this guys homework and you'll have to stay after school :)
  0
  
  Commented:
  Hehehe, someone obviously hasn't been paying attention during his programming classes.
  0
  
  Commented:
  Class project folks - I set it - beware!
  0
  
  Commented:
  Well you can do this .. it is shorter
  Question (A)
  printing from top-left to bottom right
  for index = 1 to MAX do
  print(numbers[index][index])
  end for
  printing from bottom-left to top-right
  for diag=1 to MAX do
  print(numbers[MAX-diag+1][diag])
  enfor
  
  Question (B)
  index2=0
  for index=1 to MAX
  sum=sum+(numbers[index][index2]*numbers[index][index2])
  index=index+2
  index2++
  endfor
  
  Question (C)
  I think the answer above is quite ok to follow. Most importantly must understand the concept
  0
  
  Author Commented:
  i am sorry, sir. i just want to have more opinion and more alternatives to do one programming work. i'll promise that i will not copy the direct answer
  0
  
  Commented:
  samuel, your teacher gotcha right at the place of crime :)
  
  0
  
  Commented:
  
  int sum = 0;
  
  for (int i = 0; i < 10; i++)
  sum += array[i][i];
  
  b)the square of the elements in the even rows(i.e., rows 2,4,6,8,10)
  
  int sum = 0;
  
  for (int j = 2; j < 10; j += 2) {
  for (int i = 0; i< 10; i++)
  sum += sqr(array[j][i]);
  }
  
  c)the square root of the elements in the odd columns(i.e.,columns 1,3,5,7,9).
  
  int sum = 0;
  
  for (int j = 1; j < 10; j += 2) {
  for (int i = 0; i< 10; i++)
  sum += sqrt(array[j][i]);
  }
  
  Hope this helps.
  
  -Viktor
  --Ivanov
  0
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