UVA 12716 GCD XOR

12716 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.
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

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int maxn=30000030;int d[maxn];void init(){for(int c=1;c<=maxn/2;c++){for(int a=c+c;a<maxn;a+=c){int b=a-c;if((a^b)==c) d[a]++;}}for(int i=2;i<maxn;i++)d[i]+=d[i-1];}int main(){init();int T_T,cas=1;scanf("%d",&T_T);while(T_T--){int n;scanf("%d",&n);printf("Case %d: %d\n",cas++,d[n]);}return 0;}