Even three school motto XOR problem solution

XOR
(xor.cpp / .c)
[Title] Description
Given a positive integer n, in [1, n] range, is obtained on the number of random numbers (a, b) satisfying gcd (a, b ) = a xor b. [Input format input common line, a positive integer n. [] Output format output common line, a positive integer answers. Sample [O] [input] Sample 3 [output] Sample 1 [explain] sample only (2, 3) meet the requirements. [Data] range of 30% of the data, n≤1000. For 60% of the data, n≤10 ^ 5. To 100% of the data, n≤10 ^ 7.
 

 

Title effect: Given n, find [1, n] in the number of gcd (a, b) == a ^ b; (unordered);

Assume that a> b, we know that gcd (a, b) = gcd (ab, b), while two positive integers gcd must not exceed two numbers, so gcd (a, b) ≤ab

Let a i-th bit of x, the i-th bit b to y.

When x = 1 or x = y = 0, there xy = x xor y.

When x = 0 and y = 1, there are x xor y = 1, xy = -1, i.e. x xor y> xy.

So, we have come a xor b≥ab.

Therefore, gcd (a, b) = a ^ b = ab;

Setting c = ab, then there gcd (a, ac) = c, i.e., we need to satisfy a multiple of a and c a ≠ c

Complexity: in [1, n] are multiples of the i [n / i] a, so complexity: nlogn;

#include <bits/stdc++.h>
using namespace std;
int f[10000001];
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=5000000;i++){
        for(int j=i*2;j<10000000;j+=i){
            int b=j-i;
            if((j^b)==i) f[j]++;
        }
    }
    for(int i=1;i<=n;i++) f[i]+=f[i-1];
    cout<<f[n];
}