For a given positive number we need to divide a number by 3 without using any of these *, /, +, – % operators
Examples:
Input : 48 Output : 16 Input : 16 Output : 5
Algorithm
- Take a number num, sum = 0
- while(num>3), shift the number left by 2 bits and sum = add(num >> 2, sum). Create a function add(x, y) for adding two number by bitwise operator
- Take x, y
- run a do while loop till a is not zero
- a = x & y, b = x ^ y;, y = b;
- after terminating loop return value of b which gives sum of x and y
- Take bitwise AND of num with 3
for new num = add( num >> 2, num & 3) - After terminating while loop print the value of sum which give final result .
For example if your number is 10 then convert to binary 10 => 00001010
If 10 > 3 then shift the binary number 2 bits, Now num will be 00000010 i.e, 2
Now sum will be 2
num = (shift the number by 2 bits) + (number BITWISE AND 3)
num = 2+2
Now the number will be 4 then, 4 > 3 => true so loop will be repeated
4 => 00000100 then shift the binary number 2 bits
Now sum = 2 + 1 ie, sum = 3 (this value is returned)
num = (shift the number(00000100) by 2 bits) + (number BITWISE AND 3)
num = 1 + 0
ie remainder = 1
CPP
// C++ program for divided a number by// 3 without using operator#include <bits/stdc++.h>using namespace std;// A function to add 2 numbers without using +int add(int x, int y){ int a, b; do { a = x & y; b = x ^ y; x = a << 1; y = b; } while (a); return b;}// Function to divide by 3int divideby3(int num){ int sum = 0; while (num > 3) { sum = add(num >> 2, sum); num = add(num >> 2, num & 3); } if (num == 3) sum = add(sum, 1); return sum;}// Driver programint main(void){ int num = 48; cout << "The number divided by 3 is "; cout << divideby3(num); return 0;} |
The number divided by 3 is 16
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