Question

How many + operations occur when the following pseudocode is executed? What is the output? s ← 0, t ← 0 for i ∈ {1, ..., 6} do for j ∈ {1, ..., 6} do for k ∈ {1, ..., 6} do if i + j + k = 9 then t ← t + 1 s ← s + 1 print t / s

Answer 1

code in C language corresponding to above pseudo code is:

case1

#include <stdio.h>

int main(void) {
   int s=0, t=0;
   int countOfPlus=0;
   int i,j,k;
   for(i=1;i<=6;i++)
   {
   for(j=1;j<=6;j++)
   {
   for(k=1;k<=6;k++)
   {
   countOfPlus=countOfPlus+2; // checking of if condition every time
   if(i+j+k ==9)
   {
   t=t+1;
   countOfPlus++; // when incrementing t
   s=s+1;
   countOfPlus++; // when incrementing s
   }
   countOfPlus++; //loop increment of k
   }
   countOfPlus++; //loop increment of j
   }
   countOfPlus++; //loop increment of i
   }
   printf("%d\n",t/s);
   printf("%d",countOfPlus);
   return 0;
}

count of + operation is 740

output is 1

case 2:

#include <stdio.h>

int main(void) {
   int s=0, t=0;
   int countOfPlus=0;
   int i,j,k;
   for(i=1;i<=6;i++)
   {
   for(j=1;j<=6;j++)
   {
   for(k=1;k<=6;k++)
   {
   countOfPlus=countOfPlus+2; // checking of if condition every time
   if(i+j+k ==9)
   {
   t=t+1;
   countOfPlus++; // when incrementing t
   s=s+1;
   countOfPlus++; // when incrementing s

  printf("%d\n",t/s);
   }
   countOfPlus++; //loop increment of k
   }
   countOfPlus++; //loop increment of j
   }
   countOfPlus++; //loop increment of i
}
   printf("%d",countOfPlus);
   return 0;
}

count of + operation is 740

output is 1111111111111111111111111 i.e 25 times it will print 1 because it will enter if condition for 25 times

Count of + operations. How I calculated? Simply took a variable and increased it as mentioned in comments in code. Its simple right !

Since your pseudo code was written in 1 line with no indentations, So i was confused where this "print" statement be placed?

So I made these 2 cases. Whichever suits you , take that.

I guess case 2 is what you are looking for.

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