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Day29_BitwiseAnd.cpp
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Day29_BitwiseAnd.cpp
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// THE PROBLEM
// ***************************
// Given set S={1,2,3...N}. Find two integers, A and B (where A<B), from set S
// such that the value of A&B is the maximum possible and also less than a given integer, K. In this case, & represents the bitwise AND operator.
// Solution Created By: Dustin Kaban
// Date: June 24th, 2020
// ***************************
#include <bits/stdc++.h>
using namespace std;
vector<string> split_string(string);
void max_possible_value(int n , int k)
{
std::cout << ((((k - 1) | k) > n && k % 2 == 0) ? k - 2 : k - 1) << std::endl;
}
int main()
{
int t;
cin >> t;
cin.ignore(numeric_limits<streamsize>::max(), '\n');
for (int t_itr = 0; t_itr < t; t_itr++) {
string nk_temp;
getline(cin, nk_temp);
vector<string> nk = split_string(nk_temp);
int n = stoi(nk[0]);
int k = stoi(nk[1]);
max_possible_value(n,k);
}
return 0;
}
vector<string> split_string(string input_string) {
string::iterator new_end = unique(input_string.begin(), input_string.end(), [] (const char &x, const char &y) {
return x == y and x == ' ';
});
input_string.erase(new_end, input_string.end());
while (input_string[input_string.length() - 1] == ' ') {
input_string.pop_back();
}
vector<string> splits;
char delimiter = ' ';
size_t i = 0;
size_t pos = input_string.find(delimiter);
while (pos != string::npos) {
splits.push_back(input_string.substr(i, pos - i));
i = pos + 1;
pos = input_string.find(delimiter, i);
}
splits.push_back(input_string.substr(i, min(pos, input_string.length()) - i + 1));
return splits;
}