Beer Problem Time Limit: 2 Seconds Memory Limit: 32768 KB Everyone knows that World Finals of ACM ICPC 2004 were held in Prague. Besides its greatest architecture and culture, Prague is world famous for its beer. Though drinking too much is prob…
Beer Problem Time Limit: 2000ms Memory Limit: 32768KB This problem will be judged on ZJU. Original ID: 336264-bit integer IO format: %lld Java class name: Main Everyone knows that World Finals of ACM ICPC 2004 were held in Prague. Besides its…
Factorial Problem in Base K Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3621 Description How many zeros are there in the end of s! if both s and s! are written in base k which is not nece…
比赛链接: http://acm.hust.edu.cn/vjudge/contest/view.action?cid=44704#overview 题目来源: ZOJ Monthly, June 2012 on June 24 Wine93有话说: 无话可说~~~ 转下wuyiqi巨巨的题解:http://www.cnblogs.com/wuyiqi/archive/2012/06/25/2562806.html ID Origin Title 6 / 27 Problem A Z…
题目链接 gym101778 Problem A 转化成绝对值之后算一下概率.这个题有点像 2018 ZOJ Monthly March Problem D ? 不过那个题要难一些~ #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) #define dec(i, a, b) for (int i(a); i >= (b); --i) #d…
A Simple Tree Problem Time Limit: 3 Seconds Memory Limit: 65536 KB Given a rooted tree, each node has a boolean (0 or 1) labeled on it. Initially, all the labels are 0. We define this kind of operation: given a subtree, negate all its labels. An…
这道题目还是简单的,但是自己WA了好几次,总结下: 1.对输入的总结,加上上次ZOJ Problem Set - 1334 Basically Speaking ac代码及总结这道题目的总结 题目要求输入的格式: START X Y Z END 这算做一个data set,这样反复,直到遇到ENDINPUT.我们可以先吸纳一个字符串判断其是否为ENDINPUT,若不是进入,获得XYZ后,吸纳END,再进行输出结果 2.注意题目是一个圆周,所以始终用锐角进行计算,即z=360-z; 3.知识点的误…
ZOJ Problem Set - 1025 题目分类:基础题 原题地址:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1025 题目大意就是有很多木头,都有各自的长度和重量.现在要加工这些木头,如果加工某根木头的长度和重量大于等于它上一根木头的长度和重量,那么加工它不需要时 间,否则要花1分钟.现给出一堆木头的长度和重量,要求加工完这堆木头可以花的最少时间.例如给出5根木头长度重量分别为(4,9), (5,2),…
称号:ZOJ Problem Set - 2563 Long Dominoes 题意:给出1*3的小矩形.求覆盖m*n的矩阵的最多的不同的方法数? 分析:有一道题目是1 * 2的.比較火.链接:这里 这个差点儿相同,就是当前行的状态对上一行有影响.对上上一行也有影响.所以 定义状态:dp[i][now][up]表示在第 i 行状态为now .上一行状态为 up 时的方案数. 然后转移方程:dp[i][now][up] = sum ( dp[i-1][up][uup] ) 前提是合法 合法性的推断…
ZOJ Problem Set - 3593 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3593 One Person Game Time Limit: 2 Seconds Memory Limit: 65536 KB There is an interesting and simple one person game. Suppose there is a number axis under your f…
详解OJ(Online Judge)中PHP代码的提交方法及要点 Introduction of How to submit PHP code to Online Judge Systems Introduction of How to commit submission in PHP to Online Judge Systems 在目前常用的在线oj中,codeforces.spoj.uva.zoj 等的题目可使用PHP实现基本算法,zoj是目前对PHP支持较好的中文OJ. PHP是一门比…
Prime Ring Problem Time Limit: 10 Seconds Memory Limit: 32768 KB A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime…
题目链接 ZOJ Monthly, March 2018 Problem G 题意 给定一个字符串.现在求一个下标范围$[0, n - 1]$的$01$序列$f$.$f[x] = 1$表示存在一种方案,删掉原字符串中的连续$x$个字母, 使得剩下的字符串中任意相邻的两个字母都不同.在这道题中所有的字符串首尾字符看做是相邻的. 对于每个起始位置求出最多往右延伸到的位置,满足该区间代表的字符串是一个满足任意相邻字母不同的字符串. 首先考虑一个连续的满足任意相邻字母不同的字符串.设其长度为$l$…
题目:ZOJ Problem Set - 2297 Survival 题意:给出一些怪,有两个值,打他花费的血和能够添加的血,然后有一个boss,必须把小怪全部都打死之后才干打boss,血量小于0会死.也不能大于100. 分析:定义状态:dp[st].表示在 st 状态下的血量. 然后转移:dp[st] = max (dp[st].dp[st&~(1<<i )]+p[i].first - p[i].second); 注意初始化的时候必须在開始初始化,否则easy出错. #include…
题目:problemId=5374" target="_blank">ZOJ Problem Set - 3820 Building Fire Stations 题意:给出n个点,n-1条边的一棵树.然后要在两个点上建立两个消防站.让全部点的到消防站最大距离的点的这个距离最小. 分析:首先先求这个树的直径.然后在树的直径的中点处把树分成两棵树.然后在把两棵树分别取中点的最大值就是ans值. 这个题目数据有点水了感觉... AC代码: #include <cstdi…
题目:problemId=3442" target="_blank">ZOJ Problem Set - 3229 Shoot the Bullet 分类:有源有汇有上下界网络流 题意:有 n 天和 m 个girls,然后每天给一部分girls拍照,每一个girls 有拍照的下限.即最少要拍这么多张.然后每天有k个女孩拍照,摄影师最多能够拍num张,然后 k 个女该每天拍照数量值有上下限,然后问你有没有满足这样条件的给女孩拍照的最慷慨案.然后依照输入输出每天给女孩拍照的…
