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pairs (i,j) s.t. i^j = k
By zeref_dragoneel, history, 10 years ago, In English

number of pairs (i,j) s.t. 1<=i<=n, 1<=j<=m s.t. i^j = k in O(max(log n, log m)). I am able to do it in min(n,m).

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10 years ago, hide # |
 
Vote: I like it +23 Vote: I do not like it

I will be very puzzled if that problem has a solution with logarithmic running time. And I hope that is sufficient. But if the prime factorization of k is known maybe I can solve it in .

Consider the prime factorization of k: k = p1k1p2k2... prkr , and we want to find a number of representations looks like ab = (p1a1p2a2... prar)b = p1k1p2k2... prkr .

Let g = gcd(k1, k2, ... , kr) . All of valid representations looks like (p1k1 / dp2k2 / d... prkr / d)d , where d is any divisor of g.

To satisfy 1 ≤ a ≤ n constraint we find min d such that (p1k1 / dp2k2 / d... prkr / d) ≤ n through binary search. It`s minimal valid d. Constraint 1 ≤ b ≤ m gives us maximal valid value of d. And answer is dmax - dmin + 1 .

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10 years ago, hide # |
 
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Can you please share the problem link?

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10 years ago, hide # |
 
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We need to find number of pairs (A, B) such that their XOR is K.
Consider binary representations of K — if some bit is 1 then respective bits in A and B must be different, and if bit is 0 then they need to be same.
Now we can represent N and M in binary and then solve it with DP[prefix][flag1][flag2].
prefix — how many bits are already placed in A and B.
flag1 — 0 if current prefix of A is same as prefix of N and 1 if it is less. flag2 is same as flag1 but for B.

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