Choose the correct operators to fill in the blanks:

int i,j,k; i=1;j=2;k=3; printf("%d",i___5___j___2____k);

Output is: $2$

precedence *,/,% L->R +,- L->R

i+1-2+k

= 2-2+k

= 0+k = 3

5/j-2+k = 2-2+k = 0+k = 3

i+1+2/k = i+1+0 = 2

i*5 %2 -2/k =5%2 - 2/k = 1 - 2/3

= 1 - 0

= 1 So, answer is option C .