LightOJ - 1148 - Mad Counting】的更多相关文章

lightoj 1148 Mad Counting 链接:http://lightoj.com/volume_showproblem.php?problem=1148 题意:民意调查,每一名公民都有盟友,问最少人数. 思路:考察的知识点有两个:第一是整数相乘取上整:第二是容器大小(ps:不能算一个知识点,只能算一个坑点). 做题思路:排序,如果被调查的人有相同盟友人数(n)的个数(cnt)大于这个数+1,即:(cnt > n+1), 容器已满,只能新开辟一个容器来装盟友. 代码: #includ…
先上题目: 1148 - Mad Counting   PDF (English) Statistics Forum Time Limit: 0.5 second(s) Memory Limit: 32 MB Mob was hijacked by the mayor of the Town "TruthTown". Mayor wants Mob to count the total population of the town. Now the naive approach to…
1148 - Mad Counting   PDF (English) Statistics Forum Time Limit: 0.5 second(s) Memory Limit: 32 MB Mob was hijacked by the mayor of the Town "TruthTown". Mayor wants Mob to count the total population of the town. Now the naive approach to this p…
1148 - Mad Counting   PDF (English) Statistics Forum Time Limit: 0.5 second(s) Memory Limit: 32 MB Mob was hijacked by the mayor of the Town "TruthTown". Mayor wants Mob to count the total population of the town. Now the naive approach to this p…
1148 - Mad Counting PDF (English) Statistics Forum Time Limit: 0.5 second(s) Memory Limit: 32 MB Mob was hijacked by the mayor of the Town "TruthTown". Mayor wants Mob to count the total population of the town. Now the naive approach to this pro…
http://www.lightoj.com/volume_showproblem.php?problem=1058 题意:给你顶点,问能够成多少个平行四边形. 思路:开始想使用长度来扫描有多少根,但是好像坐标太大似乎不可行.其实我们可以通过找所有线段的中点的重合个数来计算有几个平行四边形,这种通过别的性质来判断几何关系的思维是解几何题的基础,当作入门? /** @Date : 2016-12-02-21.49 * @Author : Lweleth (SoungEarlf@gmail.com)…
链接: https://vjudge.net/problem/LightOJ-1058 题意: There are n distinct points in the plane, given by their integer coordinates. Find the number of parallelograms whose vertices lie on these points. In other words, find the number of 4-element subsets o…
题目链接: http://www.lightoj.com/volume_showproblem.php?problem=1170 题目描述: 给出一些满足完美性质的一列数(x > 1 and y > 1 such that m = xy.) 然后给出一个区间,问在这个区间中的完美数组成的搜索二叉树的个数是多少?解题思路: 1,打标算出所有的完美数列中的数字 2,打表算出卡特兰数列,等着以后用 3,卡特兰数列递推式:F[N] = F[N-1] * ( 4 * N - 2 ) / ( N + 1…
链接: https://vjudge.net/problem/LightOJ-1148 题意: Mob was hijacked by the mayor of the Town "TruthTown". Mayor wants Mob to count the total population of the town. Now the naive approach to this problem will be counting people one by one. But as w…
链接: https://vjudge.net/problem/LightOJ-1170 题意: BST is the acronym for Binary Search Tree. A BST is a tree data structure with the following properties. i) Each BST contains a root node and the root may have zero, one or two children. Each of the chi…
题目链接:http://lightoj.com/volume_showproblem.php?problem=1148 已知有n个人,并且每个人都知道除自己之外还有m个人与自己支持的队一样,让我们求至少有多少个人: 就是很任性的过了: http://lightoj.com/volume_showproblem.php?problem=1148…
Text Aeroplanes are slowly driving me mad. I live near an airport and passing planes can be heard night and day. The airport was built years ago, but for some reason it could not be used then. Last year, however, it came into use. Over a hundred peop…
在上篇,我了解了基数的基本概念,现在进入Linear Counting算法的学习. 理解颇浅,还请大神指点! http://blog.codinglabs.org/articles/algorithms-for-cardinality-estimation-part-ii.html 它的基本处理方法和上篇中用bitmap统计的方法类似,但是最后要用到一个公式: 说明:m为bitmap总位数,u为0的个数,最后的结果为n的一个估计,且为最大似然估计(MLE). 那么问题来了,最大似然估计是什么东东…
来之不易的2017第一发ac http://poj.org/problem?id=2386 Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 31474   Accepted: 15724 Description Due to recent rains, water has pooled in various places in Farmer John's field, which is repr…
ZOJ3944 People Counting ZOJ3939 The Lucky Week 1.PeopleConting 题意:照片上有很多个人,用矩阵里的字符表示.一个人如下: .O. /|\ (.) 占3*3格子,句号“.”为背景.没有两个人完全重合.有的人被挡住了一部分.问照片上有几个人. 题解: 先弄个常量把3*3人形存起来,然后6个部位依次找,比如现在找头,找到一个头,就把这个人删掉(找这个人的各个部位,如果在该部位位置的不是这个人的身体,就不删),删成句号,疯狂找就行了. 代码:…
#include <stdio.h> #include <malloc.h> #define MAX_STACK 10 ; // define the node of stack typedef struct node { int data; node *next; }*Snode; // define the stack typedef struct Stack{ Snode top; Snode bottom; }*NewStack; void creatStack(NewSt…
http://lightoj.com/volume_showproblem.php?problem=1422 做的第一道区间DP的题目,试水. 参考解题报告: http://www.cnblogs.com/ziyi--caolu/p/3236035.html http://blog.csdn.net/hcbbt/article/details/15478095 dp[i][j]为第i天到第j天要穿的最少衣服,考虑第i天,如果后面的[i+1, j]天的衣服不要管,那么dp[i][j] = dp[i…
问题:http://hihocoder.com/problemset/problem/1148 给定两个日期,计算这两个日期之间有多少个2月29日(包括起始日期). 思路: 1. 将问题转换成求两个日期间有几个闰年 基本公式:闰年数(endYear,startYear) = 闰年数(0, endYear) - 闰年数(0, startYear) 注意点:区间的端点值 2. 闰年的要求 不能被100整除,能被4整除 或能被400整除 3. cnt = [i / 4] 对cnt = [i / 4]取…
题目链接:http://lightoj.com/volume_showproblem.php?problem=1298 题意:给你两个数 n, p,表示一个数是由前 k 个素数组成的,共有 n 个素数,然后求这样的所有的数的欧拉和: 例如 n = 3, p=2; 前两个素数是2,3, 然后因为n=3,所以要再选一个素数组成一个数,有两种选择2*3*2=12 和 2*3*3=18 结果就是Φ(12)+Φ(18) = 10; 我们可以用dp[i][j] 表示前 j 个素数中选择 i 个的结果,Φ[n…
1004. Counting Leaves (30)   A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Each input file contains one test case. Each case starts with a line containing 0 < N < 100,…
Problem Figure 2. The Hamming distance between these two strings is 7. Mismatched symbols are colored red. Given two strings ss and tt of equal length, the Hamming distance between ss and tt, denoted dH(s,t)dH(s,t), is the number of corresponding sym…
Problem A string is simply an ordered collection of symbols selected from some alphabet and formed into a word; the length of a string is the number of symbols that it contains. An example of a length 21 DNA string (whose alphabet contains the symbol…
http://lightoj.com/volume_showproblem.php?problem=1214 这就是一道简单的大数取余. 还想还用到了同余定理: 所谓的同余,顾名思义,就是许多的数被一个数d去除,有相同的余数.d数学上的称谓为模.如a=6,b=1,d=5,则我们说a和b是模d同余的.因为他们都有相同的余数1. //// 数学上的记法为: a≡ b(mod d) 可以看出当n<d的时候,所有的n都对d同商,比如时钟上的小时数,都小于12,所以小时数都是模12的同商. 对于同余有三种…
相关代码请戳 https://coding.net/u/tiny656/p/LightOJ/git 1006 Hex-a-bonacci. 用数组模拟记录结果,注意取模 1008 Fibsieve's Fantabulous Birthday. 找规律题,左边列是1 3平方 5平方......下边行是1 2平方 4平方......,找到当前数被包夹的位置,然后处理一下位置关系,注意奇偶. 1010 Kinghts in Chessboard. 规律题,对于m,n大于2的情况下,使用交叉放置的方法…
这里 irf 两个交换机 S4 S5 S4 S4的MEmber id 为1 IRF member 1 renumber 1 S4的 irf 优先为10 irf member priority 10 链形堆叠 irf domain 10 进入 要irf的端口 int range name irf int ten-g 1/0/50 to ten-g1/0/51 暂时关闭 shutdown 进入irf port irf-port 1/1 port group int ten-g 1/0/50 port…
// uva 11401 Triangle Counting // // 题目大意: // // 求n范围内,任意选三个不同的数,能组成三角形的个数 // // 解题方法: // // 我们设三角巷的最长的长度是c(x),另外两边为y,z // 则由z + y > x得, x - y < z < x 当y = 1时,无解 // 当y = 2时,一个解,这样到y = x - 1 时 有 x - 2个 // 解,所以一共是0,1,2,3....x - 2,一共(x - 2) * (x - 1…
当我们在使用JSONKit处理数据时,直接将文件拉进项目往往会报这两个错“JSONKit   does not support Objective-C Automatic Reference Counting(ARC)”,“ARC forbids Objective-C objects in struct”,这是由于JSONKit库未更新,不支持ARC机制.我们可以参照如下步骤解决:…
有使用JSonKit的朋友,如果遇到“JSonKit does not support Objective-C Automatic Reference Counting(ARC)”这种情况,可参照如下方法: 点击项目根目录->targets->Build Phases->JSONKit.m->添加“-fno-objc-arc”字段,在运行就OK了.…
Aladdin and the Flying Carpet Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Practice LightOJ 1341 Appoint description:  System Crawler  (2016-07-08) Description It's said that Aladdin had to solve seven myst…
A - Bi-shoe and Phi-shoe Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Practice LightOJ 1370 Appoint description:  System Crawler  (2016-07-08) Description Bamboo Pole-vault is a massively popular sport in X…