Uva 12716

题意:

输入N ， 求有多少对（a，b） gcd（a，b） == a ^ b

然而  gcd（a，b） == a ^ b = c 可以得出 b = a - c；

然后就是枚举c  a 了

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>using namespace std;const int maxn = 30000000 +131;typedef long long LL;LL num[maxn];void Int() {for(int c = 1; c <= maxn/2; ++c) //用sqrt + c*c覆盖不了全部0.0  {for(int a = c+c; a <= maxn; a+=c){if(c == (a ^ (a - c)) && (a - c))num[a]++;}}for(int i = 2; i < maxn; ++i)num[i] += num[i-1];}int main() {int T;Int();cin >> T;for(int i = 1; i <= T; ++i){int n;cin >> n;printf("Case %d: %lld\n",i,num[n]);}return 0;}