标签：style blog color for div new
1. 改变符号：取反+1
2. 与0异或保持不变，与-1(0xffffffff)异或相当于取反。
3. 负数右移可以认为是补符号位（当然也有机器不是这样子）。负数右移31位就是-1.
1 int sign(int n) { 2 return ~n + 1; 3 } 4 5 int abs(int n) { 6 int s = n >> 31; // if n >= 0, s = 0, n ^ s = n; if n < 0, s = -1, n ^ s = ~n; 7 return (n ^ s) - s; // if n >= 0, return n - 0; if n < 0, return ~n - (-1); 8 } 9 10 int swap(int &a, int &b) { 11 a ^= b; 12 b ^= a; 13 a ^= b; 14 }
求小于等于n的素数；
1 void set(int &n, int p) { 2 n |= 1 << p; 3 } 4 5 bool check(int n, int p) { 6 return n & (1 << p); 7 } 8 9 void getPrimes(int n, vector<int> &ret) { 10 int* visited = new int[n >> 5 + 1]; 11 memset(visited, 0, sizeof(int) * (n >> 5 + 1)); 12 13 for (int i = 2; i <= n; ++i) { 14 if (!check(visited[i >> 5], i & 0x1f)) { 15 ret.push_back(i); 16 for (int j = i; j <= n; j += i) { 17 set(visited[j >> 5], j & 0x1f); 18 } 19 } 20 } 21 22 delete[] visited; 23 }
反转二进制串:
1 int reverseBinary(int n) { 2 n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1); 3 n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2); 4 n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4); 5 n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8); 6 n = ((n & 0xffff0000) >> 16) | ((n & 0x0000ffff) << 16); 7 return n; 8 }
计数1的个数：
1 int calculateOnes(int n) { 2 n = ((n & 0xaaaaaaaa) >> 1) + (n & 0x55555555); 3 n = ((n & 0xcccccccc) >> 2) + (n & 0x33333333); 4 n = ((n & 0xf0f0f0f0) >> 4) + (n & 0x0f0f0f0f); 5 n = ((n & 0xff00ff00) >> 8) + (n & 0x00ff00ff); 6 n = ((n & 0xffff0000) >> 16) + (n & 0x0000ffff); 7 return n; 8 }
标签：style blog color for div new
原文地址：http://www.cnblogs.com/linyx/p/3832315.html
