GCD XOR
Given an integer N, nd 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 rst 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 rst 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.
Explanation
Sample 1: For N = 7, there are four valid pairs: (3, 2), (5, 4), (6, 4) and (7, 6).
Sample Input
2
7
20000000
Sample Output
Case 1: 4
Case 2: 34866117
题意：给出n，求出a<=n,b<=n，且gcd（a，b）==a xor b 的对数
思路：找规律得出的解，数学证明暂无，以后如果想出再补（八成补不出来= =）。找规律得出，如果a，b(a<b)满足上述条件，那么a'=a/gcd(a,b)，b'=b/gcd(a,b)，则a'=2k,b'=2k+1(k为正整数)。那么只要枚举最大公约数t，然后构造(mt)和(m+1)t，然后测试(mt)xor(m+1)t==t就行了。
/*
* Author: Joshua
* Created Time: 2014年10月05日 星期日 20时36分26秒
* File Name: h.cpp
*/
#include<cstdio>
#define maxn 30000001
int f[maxn];
int n,T; void solve()
{
for (int i=;i<maxn;++i)
for (int j=i+i;j<maxn;j+=i)
if ((j^(j-i))==i) f[j]++;
for (int i=;i<maxn;++i)
f[i]+=f[i-];
} int main()
{
solve();
scanf("%d",&T);
int x;
for (int i=;i<=T;++i)
{
scanf("%d",&x);
printf("Case %d: %d\n",i,f[x]);
}
return ;
}