题目链接
题意：在3e7的范围里有多少对gcd(a, b) = a^b。
思路:首先，我们需要知道一些前置知识.
1.a^ b = c.a^c = b;
2. a^b = c, gcd(a,b) = c, a = k * c, b = k * c;
我们可以得到gcd(a, a^c) = c。所以我们可以枚局c的所有倍数，验证gcd(a, a ^ c) == c。这样复杂度是n* log(n)*log(n)
当我们打表找规律会发现a-b=c。因为由首先有个明显的规律是a-b<=a ^ b，其次因为a = k * c, b = k * c，所以a-b>= c。因此，我们得到a-b=c这个结论。
这样我们就不用验证gcd(a, a^c) == c。只需验证 a-a ^ c == c 就行了。

#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long ll;
const int maxn = 3e7+5;
const int mod = 1e9+9;
int Case = 1, n, m, T = 0;
ll sum[maxn];
void solve() {
	scanf("%d", &n);
	printf("Case %d: %lld\n", ++T, sum[n]);
}

int main() {
    //ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
    //freopen("/Users/hannibal_lecter/Desktop/code/in.txt", "r", stdin);
    //freopen("/Users/hannibal_lecter/Desktop/code/out.txt","w",stdout);
#endif
	for(int c = 1; c <= (maxn>>1); c++) {
		for(int a = 2*c; a < maxn; a += c) {
			if(a-a^c==c) sum[a]++;
		}
	}
	for(int i = 1; i < maxn; i++) sum[i] += sum[i-1];
	scanf("%d", &Case);
    while(Case--) {
        solve();
    }
    return 0;
}

放一个会T的代码

#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long ll;
const int maxn = 3e7+5;
const int mod = 1e9+9;
int Case = 1, n, m, T = 0;
ll sum[maxn];
void solve() {
	scanf("%d", &n);
	printf("Case %d: %lld\n", ++T, sum[n]);
}

int main() {
    //ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
    //freopen("/Users/hannibal_lecter/Desktop/code/in.txt", "r", stdin);
    //freopen("/Users/hannibal_lecter/Desktop/code/out.txt","w",stdout);
#endif
	for(int c = 1; c <= (maxn>>1); c++) {
		for(int a = 2*c; a < maxn; a += c) {
			if(__gcd(a, a^c) == c) sum[a]++;
		}
	}
	for(int i = 1; i < maxn; i++) sum[i] += sum[i-1];
	scanf("%d", &Case);
    while(Case--) {
        solve();
    }
    return 0;
}