Uva12716：GCD XOR

Given an integer N, find how many pairs (A, B) are there such that: gcd(A, B) = A xor B where
1 ≤ B ≤ A ≤ N.
Here gcd(A, B) means the greatest common divisor of the numbers A and B. And A xor B is the
value of the bitwise xor operation on the binary representation of A and B.

Input

The first line of the input contains an integer T (T ≤ 10000) denoting the number of test cases. The
following T lines contain an integer N (1 ≤ N ≤ 30000000).

Output

For each test case, print the case number first in the format, ‘Case X:’ (here, X is the serial of the
input) followed by a space and then the answer for that case. There is no new-line between cases.

Sample Input

2
7
20000000

Sample Output

Case 1: 4
Case 2: 34866117

Explanation

Sample 1: For N = 7, there are four valid pairs: (3, 2), (5, 4), (6, 4) and (7, 6).

【题目描述】

给你一个正整数n，问你区间[1, n]中有多少无序数对(a, b)满足$gcd(a, b) = a \oplus b$。

【简要题解】

假设$a \gt b$，我们可以观察到

$$a \oplus b >= a - b$$
$$gcd(a,b) \le a-b$$
第一个不等式是显然的。
而由于$gcd(a,b)|b$，$gcd(a,b)|b$
所以$gcd(a,b)|a-b$，所以第二个不等式也成立
所以就有$gcd(a,b)=a-b$
我们可以枚举$c=gcd(a,b)$，再枚举$a=c*i$，判断$a \oplus c$是否等于$a-c$即可

时间复杂度：$O(nlogn)$

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int solve(int n)
{
    int ans = 0, e = n >> 1;
    for(int c = 1; c <= e; ++c)
        for(int a = c << 1; a <= n; a += c)
            if((a^c) == a-c) ans++;
    return ans;
}

int main()
{
    int n; cin >> n;
    cout << solve(n) << endl;
    return 0;
}