E - Decrease (Judge ver.) Time limit : 2sec / Memory limit : 256MB Score : 600 points Problem Statement We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest…

题目描述 We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N−1 or smaller. (The operation is the same as the one in Problem…

每次将最大的数减到n以下,如此循环直到符合题意. 复杂度大概是n*n*log?(?). #include<cstdio> #include<iostream> #include<algorithm> using namespace std; int n; typedef long long ll; ll ans,a[60]; int main(){ scanf("%d",&n); for(int i=1;i<=n;++i){ cin&g…

D - Decrease (Contestant ver.) Time limit : 2sec / Memory limit : 256MB Score : 600 points Problem Statement We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the la…

D - Decrease (Contestant ver.) Time limit : 2sec / Memory limit : 256MB Score : 600 points Problem Statement We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the la…

从n个t变化到n个t-1,恰好要n步,并且其中每一步的max值都>=t,所以把50个49当成最终局面,从这里开始,根据输入的K计算初始局面即可. #include<cstdio> #include<iostream> using namespace std; typedef long long ll; ll K; int main(){ cin>>K; int n=50; printf("%d\n",n); ll a=K/(ll)n; ll b…

构造题都是神仙题 /kk ARC066C Addition and Subtraction Hard 首先要发现两个性质: 加号右边不会有括号:显然,有括号也可以被删去,答案不变. \(op_i\)和\(A_{i+1}\)之间只会有一个括号:有多个括号的话只保留最外边那个,答案不变. 然后就可以定义状态:\(dp_{i,j}\)表示前\(i\)个数,还有\(j\)个未闭合的左括号,得到的最大答案. 由于只有减号右边有括号,所以只要知道左边有几个未闭合的左括号,就可以知道自己的贡献是\(1\)还是…

D. Decrease (Contestant ver.) 大意: 每次操作选一个最大数$-n$,其余数全$+1$. 要求构造一个序列$a$, 使得恰好$k$次操作后最大值不超过$n-1$. 只要让$k$次操作以后恰好变全为$n-1$即可. #include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include…

ARC079题解 C - Cat Snuke and a Voyage #include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define pdi pair<db,int> #define mp make_pair #define pb push_back #define enter putchar('\n') #define space putcha…

You are given a list of cities. Each direct connection between two cities has its transportation cost (an integer bigger than 0). The goal is to find the paths of minimum cost between pairs of cities. Assume that the cost of each path (which is the s…

A - ABCxxx 题意: 给出n,输出“ABCn”就可以了,纯水题. B - Break Number 题意: 给出n,找出从1到n的闭区间内能够被2整除最多次的数. 思路: 直接模拟. 代码: #include <stdio.h> #include <string.h> int main() { int n; scanf("%d",&n); ; ; ;i <= n;i++) { int tmp = i; ; == ) { tmp /= ; c…

A - ABCxxx Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement This contest, AtCoder Beginner Contest, is abbreviated as ABC. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example,…

典中典比赛 . 目录 A - Six Characters B - At Most 3 (Judge ver.) C - Poem Online Judge D - At Most 3 (Contestant ver.) E - Takahashi and Animals F - Two Spanning Trees G - Intersection of Polygons Ex - Fill Triangle A - Six Characters 题面 给一个长度不大于 \(3\) 的字符串…

前言: 自从我2014年下半年到现在的某电商公司工作后,工作太忙,一直没有写过一篇博客,甚至连14年股票市场的牛市都错过了,现在马上要过年了,而今天又是立春节气,如果再不动手,那么明年这个无春的年,也就不适合写博客了,呵呵,这仅仅对我而言.一年之计在于春,我得重新开始写一些博客,想写的很多,那么这开篇为了一个好兆头,就写PDF.NET SOD Ver 5.1完全开源 的事情吧,顺便送大家一个春节抢票工具,这两天迎来了退票高峰,没有抢到票的朋友可以试试这个工具,捡遗拾漏,顺利回家. 框架简介: “…

Atitit vod ver 12 new feature v12 pb2 影吧 视频 电影 点播 播放系统v12新特性 项目分离从独立的se ver Run mode from brow ex to self web ..for handy to dbg .. Kiosk模式支持 Ajax是最新dsl解析器 重构部分cms api分离. 要不要使用独立的api ,还是使用cms wordpress的api把.. 三大api ,cate list, single post ,post intro…

题目链接 http://codeforces.com/gym/101102/problem/C problem description Judge Bahosain was bored at ACM AmrahCPC 2016 as the winner of the contest had the first rank from the second hour until the end of the contest. Bahosain is studying the results of t…

PDF.NET 开发框架之 SOD框架 Ver 5.2.1.0307 正式版发布,包含以下部分: SOD_Pwmis.Core --包括下列数据提供程序 SqlServer SqlServerCe Access OleDb ODBC Oracle --包含框架的核心类库 PDF.NET SOD All --包括框架的全部类库和数据提供程序,目前有 SQLite MySQL PostgreSQL PDF.NET SOD AllSource --包括全部类库源码和示例程序源码,包括超市管理系统源码…

FFMpeg ver 20160219-git-98a0053 滤镜中英文对照 2016.02.21 by 1CM T.. = Timeline support 支持时间轴 .S. = Slice threading 分段线程 ..C = Command support 支持命令传送 A = Audio input/output 音频 输入/输出 V = Video input/output 视频 输入/输出 N = Dynamic number and/or type of input/out…

1 FFMpeg ver 20160213-git-588e2e3 滤镜中英文对照 2016.02.18 by 1CM 2 T.. = Timeline support 3 支持时间轴 4 .S. = Slice threading 5 分段线程 6 ..C = Command support 7 支持命令传送 8 A = Audio input/output 9 音频 输入/输出 10 V = Video input/output 11 视频 输入/输出 12 N = Dynamic numb…

Problem 1074: Hey Judge Time Limits:  1000 MS   Memory Limits:  65536 KB 64-bit interger IO format:  %lld   Java class name:  Main   Description Judge Nicole collected 7 ideas for problems of different levels, she wants to create 5 problems for the n…

转自:http://www.cnblogs.com/chouti/p/5752819.html Special Judge:当正确的输出结果不唯一的时候需要的自定义校验器 首先有个框架 #include<fstream> ifstream fin,fout,fstd ofstream fscore,freport; double Judge(){ } int main(int argc,char *argv[]) { //put something to fstreams... //Judge…

set   verify(或ver)   on/off可以设置是否显示替代变量被替代前后的语句 SQL> set verify on SQL> select &num from dual; Enter value for num: 1 old 1: select &num from dual new 1: select 1 from dual 1 ---------- 1 SQL> set verify off SQL> select &num from d…

java.lang.UnsupportedClassVersionError: com/android/dx/command/Main : Unsupported major.minor ver 解决办法: 把compile sdk version与Build tool version改成对应的. 之后在clean ,rebuild.…

前段时间准备华为机试,正好之前看了一遍<剑指 Offer>,就在九度 Online Judge 上刷了书中的题目,使用的语言为 C++:只有3题没做,其他的都做了. 正如 Linus Torvalds 所言“Talk is cheap, show me the code!",详见托管在 Github 的源码:点击打开链接(核心部分有注释). 难度总结:有些可以不必要像书中那样保存中间结果,直接输出,也就可以不使用书中的算法:有些题目在书中题目的基础上加了更多的要求,也更难实现一点.…

UVa Online Judge 工具網站   UVa中译题uHuntAlgorithmist Lucky貓的ACM園地,Lucky貓的 ACM 中譯題目 Mirror UVa Online Judge題目中譯!非常偉大的工作,請大家要心懷敬意! Ruby兔的ACM園地 UVa Online Judge題目中譯!非常偉大的工作,請大家要心懷感激! uHunt 這個站製作了一些網頁小工具, 可以查詢自己在UVa Online Judge的解題進度.簡單題列表.世界排名等等. 另外也整理了一套題庫,…

Submissions of online judge(1021) 问题描述 An online judge is a system to test programs in programming contests. The system can compile and execute codes, and test them with pre-constructed data. Submitted code may be run with restrictions, including tim…

Problem Description Ignatius is building an Online Judge, now he has worked out all the problems except the Judge System. The system has to read data from correct output file and user's result file, then the system compare the two files. If the two f…

Big endian means the most significant byte stores first in memory. int a=0x01020304, if the cpu is big endian, data are store 01 02 03 04 in memory in increasing address. Below is a simple code to make the judgement. //execute the following cmd on th…

 Hangman Judge  In ``Hangman Judge,'' you are to write a program that judges a series of Hangman games. For each game, the answer to the puzzle is given as well as the guesses. Rules are the same as the classic game of hangman, and are given as follo…

杭州电子科技大学Online Judge 之 "确定比赛名次(ID1285)"解题报告 巧若拙(欢迎转载,但请注明出处:http://blog.csdn.net/qiaoruozhuo) Problem Description 有N个比赛队(1<=N<=500).编号依次为1,2,3,.....N进行比赛.比赛结束后.裁判委员会要将全部參赛队伍从前往后依次排名. 但如今裁判委员会不能直接获得每一个队的比赛成绩,仅仅知道每场比赛的结果.即P1赢P2,用P1.P2表示,排名时P…