UVa 12716 && UVaLive 6657 GCD XOR (数论)

题意：给定一个 n ，让你求有多少对整数 (a, b) 1 <= b <= a 且 gcd（a， b） = a ^ b。

析：设 c = a ^ b 那么 c 就是 a 的约数，那么根据异或的性质 b = a ^ c，那么我们就可以枚举 a 和 c和素数筛选一样，加上gcd， n*logn*logn。

多写几个你会发现 c = a - b，证明如下：

首先 a - b <= a ^ b，且 a - b >= c，下面等于等号，用反证法，假设存在 a - b > c，那么 c < a- b <= a ^ b，然后c = a ^ b矛盾。

然后剩下就好办了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 30000000;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn+1];

int main(){
    memset(a, 0, sizeof(a));
    int m = maxn / 2;
    for(int i = 1; i <= m; i++)
        for(int j = i * 2; j <= maxn; j += i){
            int b = j - i;
            if(i == (b ^ j))   a[j]++;
        }
    for(int i = 2; i <= maxn; i++)  a[i] += a[i-1];

    int cases = 0, T, n;   cin >> T;
    while(T--){
        scanf("%d", &n);
        printf("Case %d: %d\n", ++cases, a[n]);
    }
    return 0;
}