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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <algorithm> 6 using namespace std; 7 int ans[30000010]={0}; 8 void init() 9 { 10 int i,j; 11 for(i=1;i<30000010;i++) 12 { 13 for(j=i+i;j<30000010;j+=i) 14 if((j^(j-i))==i)ans[j]++; 15 } 16 for(i=1;i<30000010;i++)ans[i]+=ans[i-1]; 17 } 18 int main() 19 { 20 int t,i,n; 21 init(); 22 scanf("%d",&t); 23 for(i=1;i<=t;i++) 24 { 25 scanf("%d",&n); 26 printf("Case %d: %d\n",i,ans[n]); 27 } 28 }