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;

}

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UVa 12716 && UVaLive 6657 GCD XOR (数论)

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