Death to Binary? Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 1707   Accepted: 529 Description The group of Absurd Calculation Maniacs has discovered a great new way how to count. Instead of using the ordinary decadic numbers, they us…

/** 题目:Death to Binary? 链接:https://vjudge.net/contest/154246#problem/T 题意:略. 思路: 注意事项: 给的字符串存在前导0: 存在0+0 */ #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <set> #include <vector>…

/* POJ2116 Death to Binary? http://poj.org/problem?id=2116 齐肯多夫定理 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <vector> #include <queue> #include <iostream> #include <m…

思路: 除去前导0,注意两个1不能相邻(11->100),注意 0 *** 或者*** 0或者0 0情况 用string的reverse()很舒服 代码: #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<queue> #include<cmath> #include<string> #include<m…

Maya Calendar Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 64795   Accepted: 19978 Description During his last sabbatical, professor M. A. Ya made a surprising discovery about the old Maya calendar. From an old knotted message, profes…

题目链接 题意 : 一个10×15的格子,有三种颜色的球,颜色相同且在同一片内的球叫做cluster(具体解释就是,两个球颜色相同且一个球可以通过上下左右到达另一个球,则这两个球属于同一个cluster,同时cluster含有至少两个球),每次选择cluster中包含同色球最多的进行消除,每次消除完之后,上边的要往下移填满空的地方,一列上的球移动之前与之后相对位置不变,如果有空列,右边的列往左移动,每一列相对位置不变 . 思路 : 模拟......不停的递归..... ////POJ 1027…

链接: http://poj.org/problem?id=3414 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22009#problem/J Pots Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8253   Accepted: 3499   Special Judge Description You are given two pots, havin…

链接:poj 2632 题意:在n*m的房间有num个机器,它们的坐标和方向已知,现给定一些指令及机器k运行的次数, L代表机器方向向左旋转90°,R代表机器方向向右旋转90°,F表示前进,每次前进一米 若前进时机器i撞到机器j,输出"Robot i crashes into robot j " 若机器走出了n*m的房间,输出"Robot i crashes into the wall " 当出现上述情况.仅仅需输出第一次出现上述的情况 若全部指令运行完,全部机器都…

题目链接:http://poj.org/problem? id=1028 Description Standard web browsers contain features to move backward and forward among the pages recently visited. One way to implement these features is to use two stacks to keep track of the pages that can be rea…

题目链接:http://poj.org/problem?id=3087 题目大意:已知两堆牌s1和s2的初始状态, 其牌数均为c,按给定规则能将他们相互交叉组合成一堆牌s12,再将s12的最底下的c块牌归为s1,最顶的c块牌归为s2,依此循环下去. 现在输入s1和s2的初始状态 以及 预想的最终状态s12.问s1 s2经过多少次洗牌之后,最终能达到状态s12,若永远不可能相同,则输出"-1". 解题思路:照着模拟就好了,只是判断是否永远不能达到状态s12需要用map,定义map<…

链接: http://poj.org/problem?id=1068 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=27454#problem/B Parencodings Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17044   Accepted: 10199 Description Let S = s1 s2...s2n be a well-forme…

题意:给定一串字符,u表示是上坡,d表示下坡,f表示平坦的,每个有不同的花费时间,问你从开始走,最远能走到. 析:直接模拟就好了,没什么可说的,就是记下时间时要记双倍的,因为要返回来的. 代码如下: #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include…

Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3840   Accepted: 2397 Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y19…

题目链接:http://poj.org/problem?id=3087 Description A common pastime for poker players at a poker table is to shuffle stacks of chips. Shuffling chips is performed by starting with two stacks of poker chips, S1 and S2, each stack containing C chips. Each…

Description A common pastime for poker players at a poker table is to shuffle stacks of chips. Shuffling chips is performed by starting with two stacks of poker chips, S1 and S2, each stack containing C chips. Each stack may contain chips of several…

题目链接 Description You, as a member of a development team for a new spell checking program, are to write a module that will check the correctness of given words using a known dictionary of all correct words in all their forms. If the word is absent in…

Description Andy is fond of old computers. He loves everything about them and he uses emulators of old operating systems on his modern computer. Andy also likes writing programs for them. Recently he has decided to write a text editor for his favorit…

传送门:http://poj.org/problem?id=1017 Packets Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 59106 Accepted: 20072 Description A factory produces products packed in square packets of the same height h and of the sizes 1*1, 2*2, 3*3, 4*4, 5*5…

题目地址:http://poj.org/problem?id=1028 测试样例: Sample Input VISIT http://acm.ashland.edu/ VISIT http://acm.baylor.edu/acmicpc/ BACK BACK BACK FORWARD VISIT http://www.ibm.com/ BACK BACK FORWARD FORWARD FORWARD QUIT Sample Output http://acm.ashland.edu/ ht…

题目代号:POJ 1573 题目链接:http://poj.org/problem?id=1573 Language: Default Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14195   Accepted: 6827 Description A robot has been programmed to follow the instructions in its path. Instr…

Instant Complexity Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 1908   Accepted: 658 Description Analyzing the run-time complexity of algorithms is an important tool for designing efficient programs that solve a problem. An algorithm…

Skew Binary Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 10676   Accepted: 6807 Description When a number is expressed in decimal, the kth digit represents a multiple of 10k. (Digits are numbered from right to left, where the least si…

Flow Layout Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 3091   Accepted: 2148 Description A flow layout manager takes rectangular objects and places them in a rectangular window from left to right. If there isn't enough room in one r…

Description Willy the spider used to live in the chemistry laboratory of Dr. Petro. He used to wander about the lab pipes and sometimes inside empty ones. One night while he was in a pipe, he fell asleep. The next morning, Dr. Petro came to the lab.…

题意:有个R*C的格网.上面有若干个点,这些点可以连成一些直线,满足:这些点在直线上均匀排布(也就是间隔相等),直线的两段穿过网格(也就是第一个,最后一个在网格的边界附近) 求某条直线上最多的点数 题解:先排序,再任取两个水稻,算出之间的dx,dy,然后就能推出前后的水稻坐标,用如果满足路径上一直有水稻(用binarysearch),且第一个点与最后一个点在水稻外面,就是一个可行的解,维护其最大值. 注意一些判断. ac代码: #include<iostream> #include<al…

                                                                                                               Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 11462   Accepted: 5558 Description A robot has been programmed to…

只需要开一个数组,记录一下这个图形. 通过一番计算,发现最大的面积大约是2k*2k的 然后递归下去染三角形. 需要计算出左上角的坐标. 然后输出的时候需要记录一下每一行最远延伸的地方,防止行末空格过多. 然后需要用putchar #include <map> #include <cmath> #include <queue> #include <cstdio> #include <cstring> #include <iostream>…

Cow Acrobats Time Limit: 1000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u Submit Status Description Farmer John's N (1 <= N <= 50,000) cows (numbered 1..N) are planning to run away and join the circus. Their hoofed feet prevent th…

Description A factory produces products packed in square packets of the same height h and of the sizes 1*1, 2*2, 3*3, 4*4, 5*5, 6*6. These products are always delivered to customers in the square parcels of the same height h as the products have and…

题意:给定s1,s1两副扑克,顺序从下到上.依次将s2,s1的扑克一张一张混合.例如s1,ABC; s2,DEF. 则第一次混合后为DAEBFC. 然后令前半段为s1, 后半段为s2. 如果可以变换成所给出的字符串,输出变换次数即可:否则,输出-1. 这题数据太水了,我是这样判断无法变换成题目所给出的字符串的:将每一次变换后的字符串存进数组,每次变换后由于之前变换后的字符串相比较,如果有相同的,就说明会无限循环下去.这样一来,循环次数会增加很多,然而0ms过了... #include <iost…