binary-bit-pattern.cpp

#include <iostream>
#include <stdio.h>
#include <string>
#define MOD 1000000007
using namespace std;
typedef long long ll;
ll i, j, n, h, ans, x, cur_h, k;
string s;
string pattern;
ll symbol_val[305];

int main() {
	freopen("input.txt", "r", stdin);
	cin >> s;

	for (char i = '0'; i <= '9'; i++)
		pattern.push_back(i);

	for (char i = 'A'; i <= 'Z'; i++)
		pattern.push_back(i);

	for (char i = 'a'; i <= 'z'; i++)
		pattern.push_back(i);

	pattern.push_back('-');
	pattern.push_back('_');

	for (i = 0; i < 64; i++)
		symbol_val[pattern[i]] = i;

	ll ans = 1;
	for (i = 0; i < s.size(); i++) {
		ll x = symbol_val[s[i]];
		for (j = 0; j < 6; j++)
			if ((x & (1 << j)) == 0)
				ans = (ans * 3) % MOD;
	}
	cout << ans << endl;
	return 0;
}


// We can transform our word in binary notation, we can do it easily, 
// because 64 = 26. Move through the bits of this number: if bit is equal to 0, 
// then we can have 3 different optinos of this bit in our pair of words: 0&1, 1&0, 0&0, 
// else we can have only one option: 1&1. So the result will be 3nullbits, 
// where nullbits — is amount of zero bits.
// https://www.hackerearth.com/practice/basic-programming/bit-manipulation/basics-of-bit-manipulation/practice-problems/algorithm/and-and-5c7e1ce5/editorial/