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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;}
UVa 12716 && UVaLive 6657 GCD XOR (数论)
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