swap(a, b): a ^= b b ^= a a ^= b 先明确一下,a ^ a = 0,同时对于一切数x ^ 0 = x 可以这样理解,第三行: b ^= a b ^= a ^ b b = b ^ a ^ b b = 0 ^ a b = a 第四行: a ^= b a = a ^ b a = (原来的a ^ b) ^ 原来的a a = b…
备份文件时看到的.我以前居然下过这东西. 2016-12-4 12:05:52更新 纯文本格式真棒.假如使用word写的我能拷过来格式还不乱?? Markdown真好. Bit Hacks By Sean Eron Anderson seander@cs.stanford.edu Converted to Markdown by Joe Gibson (@gibsjose) joseph.gibson@nasa.gov Edits and Table of Contents by Jeroen…
第三十八题 What is the bug in the following program? #include <stdlib.h> #include <stdio.h> #define SIZE 15 int main() { int *a, i; a = malloc(SIZE*sizeof(int)); ; i<SIZE; i++) *(a + i) = i * i; ; i<SIZE; i++) printf("%d\n", *a++);…
http://graphics.stanford.edu/~seander/bithacks.html Bit Twiddling Hacks By Sean Eron Andersonseander@cs.stanford.edu Individually, the code snippets here are in the public domain (unless otherwise noted) — feel free to use them however you please. Th…
题意:一棵树上有n个节点,编号分别为1到n,每个节点都有一个权值w.我们将以下面的形式来要求你对这棵树完成一些操作: I. CHANGE u t : 把结点u的权值改为t II. QMAX u v: 询问从点u到点v的路径上的节点的最大权值 I II. QSUM u v: 询问从点u到点v的路径上的节点的权值和 注意:从点u到点v的路径上的节点包括u和v本身 1<=n<=30000,0<=q<=200000:中途操作中保证每个节点的权值w在-30000到30000之间. 思路:树链…
http://acm.hdu.edu.cn/showproblem.php?pid=5661 Claris and XOR Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 146 Accepted Submission(s): 51 Problem Description Claris loves bitwise operatio…
Ba Gua Zhen Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description During the Three-Kingdom period, there was a general named Xun Lu who belonged to Kingdom Wu. Once his troop were chasing Bei Liu, he was stuck in the Ba Gua Zhen from Liang Zhuge.…
XOR Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1364 Accepted Submission(s): 402 Problem Description XOR is a kind of bit operator, we define that as follow: for two binary base number A…
