AND Product


Problem Statement :


Consider two non-negative long integers,  and , where . The bitwise AND of all long integers in the inclusive range between  and  can be expressed as , where  is the bitwise AND operator.

Given  pairs of long integers,  and , compute and print the bitwise AND of all natural numbers in the inclusive range between  and .

For example, if  and , the calculation is .

Function Description

Complete the andProduct in the editor below. It should return the computed value as an integer.

andProduct has the following parameter(s):

a: an integer
b: an integer
Input Format

The first line contains a single integer , the number of intervals to test.
Each of the next  lines contains two space-separated integers  and .

Output Format

For each pair of long integers, print the bitwise AND of all numbers in the inclusive range between  and  on a new line.



Solution :



title-img


                            Solution in C :

In C :




#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int main()
{
	int t;
	unsigned int a, b;
	
	scanf("%d", &t);
	
	while (t--) {
		scanf("%u%u", &a, &b);
		printf("%u\n", a);
	}
	
	return 0;
}
                        


                        Solution in C++ :

In  C++  :





#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ll;
ll a, b, o = 1;

bool all(ll x, ll a, ll b) {
	return (a&x) && (b&x) && (b - a) < x;
}

int main() {
	int T; scanf("%d", &T); for(int ks = 1; ks <= T; ++ks) {
		scanf("%llu%llu", &a, &b);
		ll bb = 0;
		for(int i = 0; i < 63; ++i) {
			ll t = o << i;
			if(all(t, a, b)) bb |= t;
		}
		printf("%llu\n", bb);
	}
	return 0;
}
                    


                        Solution in Java :

In  Java :







import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class B {
	static InputStream is;
	static PrintWriter out;
	static String INPUT = "";
	
	static void solve()
	{
		for(int T = ni();T >= 1;T--){
			long a = nl(), b = nl();
			long ret = 0;
			for(int i = 32;i >= 0;i--){
				if((a^b)<<~i>=0){
					ret |= a&(1L<<i);
				}else{
					break;
				}
			}
			out.println(ret);
		}
	}
	
	public static void main(String[] args) throws Exception
	{
		long S = System.currentTimeMillis();
		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		solve();
		out.flush();
		long G = System.currentTimeMillis();
		tr(G-S+"ms");
	}
	
	private static boolean eof()
	{
		if(lenbuf == -1)return true;
		int lptr = ptrbuf;
		while(lptr < lenbuf)if(!isSpaceChar(inbuf[lptr++]))return false;
		
		try {
			is.mark(1000);
			while(true){
				int b = is.read();
				if(b == -1){
					is.reset();
					return true;
				}else if(!isSpaceChar(b)){
					is.reset();
					return false;
				}
			}
		} catch (IOException e) {
			return true;
		}
	}
	
	private static byte[] inbuf = new byte[1024];
	static int lenbuf = 0, ptrbuf = 0;
	
	private static int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	
	private static boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private static int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	
	private static double nd() { return Double.parseDouble(ns()); }
	private static char nc() { return (char)skip(); }
	
	private static String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	private static char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	private static char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	private static int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	
	private static int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private static long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private static void tr(Object... o) { if(INPUT.length() != 0)System.out.println(Arrays.deepToString(o)); }
}
                    


                        Solution in Python : 
                            
In  Python3 :






T = int(input())

for _ in range(T) :
    A,B = (int(_) for _ in input().split())
    i,C = -1,A^B
    while C > 0 :
        C >>= 1
        i += 1
    print(A & (2**32-2**i))
                    


View More Similar Problems

The Strange Function

One of the most important skills a programmer needs to learn early on is the ability to pose a problem in an abstract way. This skill is important not just for researchers but also in applied fields like software engineering and web development. You are able to solve most of a problem, except for one last subproblem, which you have posed in an abstract way as follows: Given an array consisting

View Solution →

Self-Driving Bus

Treeland is a country with n cities and n - 1 roads. There is exactly one path between any two cities. The ruler of Treeland wants to implement a self-driving bus system and asks tree-loving Alex to plan the bus routes. Alex decides that each route must contain a subset of connected cities; a subset of cities is connected if the following two conditions are true: There is a path between ever

View Solution →

Unique Colors

You are given an unrooted tree of n nodes numbered from 1 to n . Each node i has a color, ci. Let d( i , j ) be the number of different colors in the path between node i and node j. For each node i, calculate the value of sum, defined as follows: Your task is to print the value of sumi for each node 1 <= i <= n. Input Format The first line contains a single integer, n, denoti

View Solution →

Fibonacci Numbers Tree

Shashank loves trees and math. He has a rooted tree, T , consisting of N nodes uniquely labeled with integers in the inclusive range [1 , N ]. The node labeled as 1 is the root node of tree , and each node in is associated with some positive integer value (all values are initially ). Let's define Fk as the Kth Fibonacci number. Shashank wants to perform 22 types of operations over his tree, T

View Solution →

Pair Sums

Given an array, we define its value to be the value obtained by following these instructions: Write down all pairs of numbers from this array. Compute the product of each pair. Find the sum of all the products. For example, for a given array, for a given array [7,2 ,-1 ,2 ] Note that ( 7 , 2 ) is listed twice, one for each occurrence of 2. Given an array of integers, find the largest v

View Solution →

Lazy White Falcon

White Falcon just solved the data structure problem below using heavy-light decomposition. Can you help her find a new solution that doesn't require implementing any fancy techniques? There are 2 types of query operations that can be performed on a tree: 1 u x: Assign x as the value of node u. 2 u v: Print the sum of the node values in the unique path from node u to node v. Given a tree wi

View Solution →