Блог пользователя CazadorDivino

bitwise, when we use this ((k >> i) & 1) == 0
Автор CazadorDivino, история, 9 лет назад, По-английски

In the last codeforces 307 div 2, problem D. One solutione is below, What that mean 

  if (((k >> i) & 1) == 0) res = res * fib % mod;
            else
                res = res * (pow2 - fib) % mod;

in this code.

long pow2 = fastPower(2, n);
        long fib = fibonacci(n + 1);
        long res = 1;
        for (int i = 0; i < l; i++)
            if (((k >> i) & 1) == 0) res = res * fib % mod;
            else
                res = res * (pow2 - fib) % mod;
        if (res < 0) res += mod;
        System.out.println(res);
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9 лет назад, # |
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The >> in (k >> i) is a shift right function on the binary representation of the number k by shifting it right for i times. In the other words, it returns the number that delete the last i digits from the binary representation of k. For examples:

(5 >> 1) returns 2 ( 510 = 1012, after deleting the last digit, it changes to 102, which gives 210. )

(42 >> 3) returns 5 ( 4210 = 1010102, after deleting the last 3 digits, it changes to 1012, which gives 510. )

Second, (x & 1) is a common way to check whether if the last digit in the binary representation of x is 1 or not. It returns 1 if the last digit in the binary representation of x is 1.

So, (((k >> i) & 1) == 0) is determining whether if the last digit in the binary representation of k after deleting the last i digits is zero or one. In the other words, it is checking if the (i + 1)th digit starting from right to the left is zero or not.

Sorry if this is long, but I hope you can understand the logic behind this code :)

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