ACM: FZU 2102 Solve equation - 手速题】的更多相关文章

 FZU 2102   Solve equation Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u  Practice Description You are given two positive integers A and B in Base C. For the equation: A=k*B+d We know there always existing many non-nega…
C Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice FZU 2102 Description You are given two positive integers A and B in Base C. For the equation: A=k*B+d We know there always existing many non-negativ…
                                                                                                                                     Problem 2102 Solve equation Accept: 1097    Submit: 2608Time Limit: 1000 mSec    Memory Limit : 32768 KB Problem Desc…
表达式求值 时间限制:1000 ms | 内存限制:65535 KB 难度:3   描述 假设表达式定义为:1. 一个十进制的正整数 X 是一个表达式.2. 如果 X 和 Y 是 表达式,则 X+Y, X*Y 也是表达式; *优先级高于+.3. 如果 X 和 Y 是 表达式,则 函数 Smax(X,Y)也是表达式,其值为:先分别求出 X ,Y值的各位数字之和,再从中选最大数.4.如果 X 是 表达式,则 (X)也是表达式. 例如:表达式 12*(2+3)+Smax(333,220+280) 的值…
 Gym 101047B  Renzo and the palindromic decoration Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Practice Description standard input/output At the ruins of Wat Phra Si Sanphet (วดพระศรสรรเพชญ), one can find famous inscr…
 FZU 2110  Star Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u  Practice Description Overpower often go to the playground with classmates. They play and chat on the playground. One day, there are a lot of stars in the sk…
解法: 1.先将对应的字符串存入map. 2.然后将输入的串的second置为空. 3.输出6-n,输出map中的非空串. 代码: #include <bits/stdc++.h> using namespace std; typedef long long ll; map <string,string> m; int main(){ m["purple"] = "Power"; m["green"] = "Ti…
题目列表: 2146 Problem A [手速]阔绰的Dim 2147 Problem B [手速]颓废的Dim 2148 Problem C [手速]我的滑板鞋 2149 Problem D [手速]潦倒的Dim 2150 Problem E [手速]被NTR的Dim 2146 Problem A: 简单的最长回文串统计算法,这里没有过高要求,n^2算法可以AC.其中包括dp动规以及中心法(以上两种都是O(n^2)算法,可以参考白书).推广,可以尝试扩展KMP(O(nlogn))或者Mana…
题目链接:fzu 1909 An Equation 典型的签到题. #include <stdio.h> #include <string.h> bool judge(int *a, int sum) { for (int i = 0; i < 4; i++) { for (int j = i + 1; j < 4; j++) if (sum == (a[i] + a[j]) * 2) return true; } return false; } int main ()…
/** 题目:Solve Equation 链接:http://acm.hnust.edu.cn/JudgeOnline/problem.php?id=1643 //最终来源neu oj 2014新生选拔赛题 题意:给定两个数的和以及他们的最小公倍数,求这两个数. 思路: x+y=A lcm(x,y)=B => x*y/gcd(x,y)=B 要把这两个公式联立,那么必须消掉gcd: 设:d = gcd(x,y), x = kx*d, y = ky*d; kx与ky互质: x+y=A => d(…
显然D是个细节题,但是还剩1h时看眼榜还没人过EF,只好冷静写D,大概思路是任何时候如果min(n,m)<=2,max(n,m)<=4暴搜,否则直接贪心是很对的,即第一步让S.T长度平均化,接下来再分01讨论,都只取头上. 然后最后1min过了样例,最后又检查了若干秒,决定大力交题,当时的时间已经只剩15s了.在这紧要关头,网卡了! QAQ 不管怎样一方面也是自己手速慢把QAQ 还挺难受的(当然如果最后那发是挂的我就会好受很多QwQ) [upd]事实上n=1,m>4时还是会有奇怪的事情,…
Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6023    Accepted Submission(s): 2846 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y…
Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 22742    Accepted Submission(s): 9865 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 ==…
蛋疼啊,本次只做出了一道题目...渣爆了... 妈蛋,,卡题之夜..比赛结果是1道题,比赛完哗啦哗啦出4道题.. A acdream1191 Dragon Maze 题意: 给一个迷宫,给出入口坐标和出口坐标,要求从入口到出口的步数尽可能少,如果有多种方案,则要求获得的分数尽可能多,获得的分数为经过的方格的数字之和 思路: bfs求最小步数,每走一步更新一下走到这个格子的最大权值 #include <bits/stdc++.h> using namespace std; typedef lon…
CSU 1772 漫漫上学路 Time Limit: 1000MS   Memory Limit: 131072KB   64bit IO Format: %lld & %llu Submit Status Description 对于csuxushu来说,能够在CSU(California State University)上学是他一生的荣幸.CSU校园内的道路设计的十分精巧,由n+1条水平道路和n+1条竖直道路等距交错而成,充分体现了校园深厚的文化底蕴.然而不幸的是CS市每到夏季,天降大雨,…
 FZU 2105  Digits Count Time Limit:10000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Practice Description Given N integers A={A[0],A[1],...,A[N-1]}. Here we have some operations: Operation 1: AND opn L R Here opn, L and R are intege…
 FZU 2112 Tickets Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Practice Description You have won a collection of tickets on luxury cruisers. Each ticket can be used only once, but can be used in either direction betwee…
Codeforces Round #262 (Div. 2) B B - Little Dima and Equation B. Little Dima and Equation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Dima misbehaved during a math lesson a lot a…
FZU 2150 Fire Game Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u   Description Fat brother and Maze are playing a kind of special (hentai) game on an N*M board (N rows, M columns). At the beginning, each grid of this bo…
POJ 1401 Factorial Time Limit:1500MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu   Description The most important part of a GSM network is so called Base Transceiver Station (BTS). These transceivers form the areas called cells (this ter…
wywcgs: 亦称Lord Wu,俗名吴垠,2009级厦门大学智能科学与技术学院研究生,本科就读于哈尔滨工业大学.因其深厚的算法功底与独到的思维方式,被尊为“吴教主”,至今声威犹存. 2006年起参加ACM/ICPC竞赛,获得分别获得上海.西安.长春三枚区域赛银牌. 2008年获得GCJ-BeiJing Onsite参赛资格,并在比赛中表现优异,获得Final资格,后因个人原因,推掉了总决赛资格,并忙于为各赛区出题. 2010年复出比赛,在福建省举办的全国邀请赛中力压群牛,获得赛区金牌,给同场…
点击打开链接 2226: Contest Print Server Time Limit: 1 Sec  Memory Limit: 128 MB Submit: 53  Solved: 18 [Submit][Status][Web Board] Description In ACM/ICPC on-site contests ,3 students share 1 computer,so you can print your source code any time. Here you ne…
版权声明:本文为博主原创文章.未经博主同意不得转载. https://blog.csdn.net/DaiHaoC83E15/article/details/26187183        回到作案现场:http://acdream.info/onecontest/1014        前言:自己出份山寨版的解题报告. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -…
 目录: K Airdrop I Soldier Game L Sub-cycle Graph G Repair the Artwork ———————————————————— ps:楼主脑残有点严重,很容易写错别字和语言组织混乱,如果在读文章时遇到,可以在评论区里回复一下,我好改(花式骗评论) 补题地址:http://acm.zju.edu.cn/onlinejudge/showProblems.do?contestId=1&pageNumber=31 顺便好人做到底,给大家凑个11/13的…
F. Palindrome Problem Description A string is palindrome if it can be read the same way in either direction, for example "maram" is palindrome, while "ammar" is not. You are given a string of n characters, where each character is eithe…
FZU 2107 Hua Rong Dao Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u  Practice  Description Cao Cao was hunted down by thousands of enemy soldiers when he escaped from Hua Rong Dao. Assuming Hua Rong Dao is a narrow aisl…
前言 demo预览->typewriter gameⅡ (chrome only 没做兼容) 别看一开始时速度不快,会线性增长的哦,反正楼主的score还没达到过40... 为什么叫Ⅱ呢?之前写了个打字机游戏(试玩猛戳这里),数据上只是简单的手写了50个单词,全部输入正确即可过关,本来想改一下做成竞技类的(有分数的),但是一样的小游戏写俩没啥意思,正好最近看了下css3,突然有了某种灵感,于是typewriter gameⅡ粗线了. 过程 之前的打字机游戏是基于canvas和粒子系统的,这次是基…
 FZU 2148  Moon Game Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u   Description Fat brother and Maze are playing a kind of special (hentai) game in the clearly blue sky which we can just consider as a kind of two-dimen…
http://acm.hdu.edu.cn/showproblem.php?pid=4455 https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4384 题目大意就不多说了,官方的解法是dp,没太理解,自己想了一个直接点的方法,O(n). 既然要计算所有贡献和,对于区间长度为k,假设集合中的元素全都相同,那么这个元素将会贡献给所…
序列变换 alpq654321 (命题人)   基准时间限制:1 秒 空间限制:131072 KB 分值: 40 lyk有两序列a和b. lyk想知道存在多少对x,y,满足以下两个条件. 1:gcd(x,y)=1. 2: a[b[x]] = b[a[y]] .   例如若a={1,1,1},b={1,1,1}.那么存在7对,因为除了x=2,y=2或x=3,y=3外都满足条件. Input 第一行一个数n(1<=n<=100000). 接下来一行n个数,表示ai(1<=ai<=n).…