题意 分析 考场做法 考虑二分答案,R开到1e9就能过了. 判断答案合法,就判断时间和是否超过拥有的时间就行了.但要把di从小到大排序,不然容易验证贪心是错的. 时间复杂度\(O(n \log n)\) #include<cstdlib> #include<cstdio> #include<cmath> #include<cstring> #include<iostream> #include<string> #include<…

Xiao Ming's Hope Time Limit:1000MS     Memory Limit:32768KB  Description Xiao Ming likes counting numbers very much, especially he is fond of counting odd numbers. Maybe he thinks it is the best way to show he is alone without a girl friend. The day…

Xiao Ming's Hope Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1668    Accepted Submission(s): 1109 Problem Description Xiao Ming likes counting numbers very much, especially he is fond of cou…

原文地址:http://iwantmoon.com/Post/487ab43d609f49d28ff4228241e2b7c7 Rpc(Remote Procedure Call Protocal)远程调用协议,分布式项目的支柱. Ming Rpc(Ming Remote Procedure Call)是为MFS(Ming File System)设计的Rpc框架,特点在于Ming Rpc将过程开放给Command,所有共同处理部分在Rpc Server中封装. 流程简述: 1.Server s…

Ming是一个操纵swf(flash movice)的C库,支持php. ruby. python等语言. 重要提示: 在安装Ming之前,应该准备好你的系统,特别是Linux/Unix系统,如果你对系统配置不是很熟悉,最好的办法是在安装系统的时候要将绝大部分的开发模块安装好. Ming的安装分为两步,首先安装Ming库,然后再安装语言相关的模块. 下载:http://sourceforge.net/projects/ming/files/Releases/ming-0.4.5.tar.gz/d…

Baby Ming and Weight lifting Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 681    Accepted Submission(s): 280 Problem Description Baby Ming is fond of weight lifting. He has a barbell pole(the…

Xiao Ming climbing Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=629&pid=1002 Description 小明因为受到大魔王的诅咒,被困到了一座荒无人烟的山上并无法脱离.这座山很奇怪: 这座山的底面是矩形的,而且矩形的每一小块都有一个特定的坐标(x,y)(x,y)和一个高度HH. 为了逃离这座…

Baby Ming and Matrix games 题意: 给一个矩形,两个0~9的数字之间隔一个数学运算符(‘+’,’-‘,’*’,’/’),其中’/’表示分数除,再给一个目标的值,问是否存在从一个数字出发,以数字之间的运算符为运算,得到这个目标值:(每个数字只能用一次,其实说白了就是dfs..);可以则输出(Impossible),否则输出(Possible); 思路:坑点就是里面本来全是整数,但是一个除法运算却是分数形式,开始使用了很保险的分数保存,来避免误差的.但是无情WA了很多次..…

Problem Description These few days, Baby Ming is addicted to playing a matrix game. Given a n∗m matrix, the character ,j∗) (i,j=,,...) are the numbers between −. There are an arithmetic sign (‘+’, ‘-‘, ‘∗’, ‘/’) between every two adjacent numbers, ot…

Problem Description Baby Ming collected lots of cell phone numbers, and he wants to sell them for money. He thinks normal number can be sold for b yuan, while number with following features can be sold for a yuan. .The last five numbers are the same.…

Problem Description   Due to the curse made by the devil,Xiao Ming is stranded on a mountain and can hardly escape. This mountain is pretty strange that its underside is a rectangle which size is n∗m and every little part has a special coordinate(x,y…

Problem Description Baby Ming is fond of weight lifting. He has a barbell pole(the weight of which can be ignored) and two different kinds of barbell disks(the weight of which are respectively a and b), the amount of each one being infinite.Baby Ming…

Baby Ming and Matrix games Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1210    Accepted Submission(s): 316 Problem Description These few days, Baby Ming is addicted to playing a matrix game…

翻译计划     小明初学者C++,它确定了四个算术.关系运算符.逻辑运算.颂值操作.输入输出.使用简单的选择和循环结构.但他的英语不是很好,记住太多的保留字,他利用汉语拼音的保留字,小屋C++,发明了一种表达自己思想的算法描写叙述规则.     规则非常easy:他将開始程序头部以一个拼音名字标记,C++程序中的"{,}"用拼音"kaishi,jieshu"直观表示.选择和循环仅仅採用一种单一的结构,且保留字也分别用相应的拼音表示,只是在表示选择或循环条件时他去掉…

Xiao Ming's Hope Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Xiao Ming likes counting numbers very much, especially he is fond of counting odd numbers. Maybe he thinks it is the best way to…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5433 Xiao Ming climbing Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1346    Accepted Submission(s): 384 Problem Description Due to the curse m…

Xiao Ming's Hope Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1786    Accepted Submission(s): 1182 Problem Description Xiao Ming likes counting numbers very much, especially he is fond of cou…

Baby Ming and Matrix games Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 849    Accepted Submission(s): 211 Problem Description These few days, Baby Ming is addicted to playing a matrix game.…

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=4349 Xiao Ming's Hope Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1723    Accepted Submission(s): 1144 Problem Description Xiao Ming likes coun…

 Baby Ming and phone number Crawling in process... Crawling failed Time Limit:1500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 5611 Description Baby Ming collected lots of cell phone numbers, and he wants…

 Baby Ming and Matrix games Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1150    Accepted Submission(s): 298 Problem Description These few days, Baby Ming is addicted to playing a matrix g…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 438    Accepted Submission(s): 108 Problem Description Due to the curse made by the devil,Xiao Ming is stranded on a mountain and can hardly esca…

这篇文章汇总了掘金前站长Ming Yin(阴明)在知乎的几个犀利的观点,原文可访问zhihu.com/kalasoo 由@flightmakers转载(收藏)在此 你是否有个人网站.可否和大家分享一下你的个人网站.你觉得个人网站在自媒体盛行的今天最大的价值是什么? 谢邀,我的个人网站网站:@kalasoo 我想,个人网站的没落已经是一个事实,用户会在社交网络和内容平台上生产内容,那里的个人主页就是你的象征.而没落背后的原因是 Open Internet 的结束,当然他的价值还很大,流畅自由的互联…

有这样一个性质:C(n,m)%p=C(p1,q1)*C(p2,q2).......%p,其中pkpk-1...p1,qkqk-1...q1分别是n,m在p进制下的组成. 就完了. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ; int main() { while (scanf("%d",&am…

题意:给一张地图,给出起点和终点,每移动一步消耗体力abs(h1 - h2) / k的体力,k为当前斗志,然后消耗1斗志,要求到终点时斗志大于0,最少消耗多少体力. 解法:bfs.可以直接bfs,用dp维护最小值……也可以用优先队列优化……但是不能找到终点后就直接输出,因为从不同方向到达终点的消耗不同,终点的前一个状态不一定比这一状态更优……一开始并没有意识到这一点……wa了一篇……后来加了dp维护……但其实在将点弹出的之后再更新vis也可以……以前因为更新vis的问题T过……留下了阴影…… 代…

输入一个n(1<=n<=108),求C(n,0),C(n,1),C(n,2)...C(n,n)有多少个奇数. Lacus定理 http://blog.csdn.net/acm_cxlove/article/details/7844973 A.B是非负整数,p是质数.AB写成p进制:A=a[n]a[n-1]...a[0],B=b[n]b[n-1]...b[0]. 则组合数C(A,B)与C(a[n],b[n])*C(a[n-1],b[n-1])*...*C(a[0],b[0])  modp同 所以…

题意:小明因为受到大魔王的诅咒,被困到了一座荒无人烟的山上并无法脱离.这座山很奇怪: 这座山的底面是矩形的,而且矩形的每一小块都有一个特定的坐标(x,y)和一个高度H. 为了逃离这座山,小明必须找到大魔王,并消灭它以消除诅咒. 小明一开始有一个斗志值k,如果斗志为0则无法与大魔王战斗,也就意味着失败. 小明每一步都能从他现在的位置走到他的(N,E,S,W)(N,E,S,W)(N,E,S,W)四个位置中的一个,会消耗(abs(H​1​​−H​2​​))/k的体力,每走一步消耗一点斗志. 大魔王很强…

题目链接 给一个n, 求C(n, 0), C(n, 1), ..........C(n, n)里面有多少个是奇数. 我们考虑lucas定理, C(n, m) %2= C(n%2, m%2)*C(n/2, m/2)%2,   C(n/2, m/2) = C(n/2%2, m/2%2)*C(n/2/2, m/2/2), 这样一直递归下去,直到m为0. 我们知道如果一个数是奇数, 那么它的所有因子都是奇数, 对应于上面的式子, n%2是偶数的时候, m%2也必须是偶数才可以, 而n%2是奇数的时候,…

非常无语的一个题. 反正我后来看题解全然不是一个道上的. 要用什么组合数学的lucas定理. 表示自己就推了前面几个数然后找找规律. C(n, m) 就是 组合n取m: (m!(n-m!)/n!) 假设n==11 : C(11,0):C(11,1):C(11,2):C(11,3):C(11,4):C(11,5): 分别为 (1/1); (1 / 11) ; (11*10 / 2*1)  ;   (11*10*9 / 3*2*1); (11*10*9*8 / 4*3*2*1) ;  (11*10*…

题目链接 题解:题意为给出一个N*M的矩阵,然后(i∗2,j∗2) (i,j=0,1,2...)的点处是数字,两个数字之间是符号,其他位置是‘#’号. 但不知道是理解的问题还是题目描述的问题,数据中还有类似1#1这种数据存在,因此WA了4次,加上了一句代码后,马上AC了,该行代码在下文以斜粗体标出. 此外,因为里面有除法,会有一定误差,所以用这句来判断:fabs(tar-nnum)<=1e-8,而不是tar==nnum.   #include <cstdio> #include <…