B. Alternating Current Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/343/problem/B Description Mad scientist Mike has just finished constructing a new device to search for extraterrestrial intelligence! He was in such a hu…
D. Alternating Current time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mad scientist Mike has just finished constructing a new device to search for extraterrestrial intelligence! He was in…
A. Magnets 模拟. B. Simple Molecules 设12.13.23边的条数,列出三个等式,解即可. C. Rational Resistance 题目每次扩展的电阻之一是1Ω的,所以假设当前的电阻为\(\frac{a}{b}\)会变成\(\frac{a+b}{b}\)或\(\frac{a}{a+b}\),其实就是求gcd的过程. D. Alternating Current 若出现连续两个相同的符号,其实线的方向是不变的,最后就是转化成括号匹配. E. Read Time…
C. Read Time Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/343/problem/C Description Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data i…
A. Rational Resistance Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/343/problem/A Description Mad scientist Mike is building a time machine in his spare time. To finish the work, he needs a resistor with a certain resista…
D. Water Tree Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/343/problem/D Description Mad scientist Mike has constructed a rooted tree, which consists of n vertices. Each vertex is a reservoir which can be either empty or…
C. Rational Resistance time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mad scientist Mike is building a time machine in his spare time. To finish the work, he needs a resistor with a certai…
为了锻炼个人能力奋力div1 为了不做原题从200开始 B 两个电线缠在一起了 能不能抓住两头一扯就给扯分开 很明显当len为odd的时候无解 当len为偶数的时候 可以任选一段长度为even的相同字符串给它翻过去 即 ++--++++--++ -> ++--------++ -> ++++++++++++ 一直这样下去一定可以翻出结果 但是复杂度很高 不能暴力的去找 考虑用一个栈 依次往里放 每次看到top等于当前的这个字符 意义是当前有一个偶数了  便pop出来 最后判断栈的empty 做…
思路: dfs序其实是很水的东西.  和树链剖分一样, 都是对树链的hash. 该题做法是:每次对子树全部赋值为1,对一个点赋值为0,查询子树最小值. 该题需要注意的是:当我们对一棵子树全都赋值为1的时候, 我们要查询一下赋值前子树最小值是不是0, 如果是的话, 要让该子树父节点变成0, 否则变0的信息会丢失. 细节参见代码: #include <cstdio> #include <cstring> #include <algorithm> #include <i…
题目链接 这题,关键不是二分,而是如果在t的时间内,将n个头,刷完这m个磁盘. 看了一下题解,完全不知怎么弄.用一个指针从pre,枚举m,讨论一下.只需考虑,每一个磁盘是从右边的头,刷过来的(左边来的之前刷了). 思维是硬伤. #include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <cmath> #include <algori…