Dont know if its valid, but imagine you have
a=5= 101;
b=2= 010;
logical and in C/C++:
5 = true
2 = true
5 && 2
<=>
true && true => true
Bitwise:
5 = 101
2 = 010
5 & 2
<=>
101 & 010 => 000
I dont remember well if && was for logical and & for bitwise or if im wrong, but i think it works this way
For the first problem you should sum up 50s until it would be too much and then repeat with 20s and 10s
Anonymous
For question K take the input
Step 1)divide input by 100 if remainder 0 then result of division is 100 denomination no. If not
Step2) divide remainder by 50 and check if remainder.
,If yes divide by 10 again until no remainder...Keep a count on each division and add.
I think that should do