UVA 12716 GCD XOR（GCD等于XOR）（数论）

题意：输入整数n(1<=n<=30000000)，有多少对整数(a, b)满足：1<=b<=a<=n，且gcd(a,b)=a XOR b。

分析：因为c是a的约数，所以枚举c，a = k*c，通过a-c求b，并通过a^b=c来验证。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 30000000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int ans[MAXN];
void init(){
    for(int c = 1; c <= (MAXN >> 1); ++c){
        for(int a = c + c; a <= MAXN; a += c){
            int b = a - c;
            if((a ^ b) == c) ++ans[a];
        }
    }
    for(int i = 2; i <= MAXN; ++i){
        ans[i] += ans[i - 1];
    }
}
int main(){
    init();
    int T;
    scanf("%d", &T);
    int kase = 0;
    while(T--){
        int n;
        scanf("%d", &n);
        printf("Case %d: %d\n", ++kase, ans[n]);
    }
    return 0;
}