Can you find it? Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)Total Submission(s): 19416    Accepted Submission(s): 4891 Problem Description Give you three sequences of numbers A, B, C, then we give you a number…

Equations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6927    Accepted Submission(s): 2810 Problem Description Consider equations having the following form: a*x1^2+b*x2^2+c*x3^2+d*x4^2=0a, b…

Pie Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7302    Accepted Submission(s): 2732 Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I h…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2141 题目大意:查找是否又满足条件的x值. 这里简单介绍一个小算法,二分查找. /* x^2+6*x-7==y 输入y 求x 精确度为10^-5 0=<x<=10000 */ #include <iostream> #include <cstdio> using namespace std; int main (void) { double y; while(cin>…

Can you find it? Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)Total Submission(s): 9180    Accepted Submission(s): 2401 Problem Description Give you three sequences of numbers A, B, C, then we give you a number…

Problem Description Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.   Input There are many cases. Every data cas…

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X. Input There are many cases. Every data case is described as foll…

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X. Input There are many cases. Every data case is described as foll…

题目链接: Can you find it? Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)Total Submission(s): 38634    Accepted Submission(s): 9395 Problem Description Give you three sequences of numbers A, B, C, then we give you a…

题目传送门 /* 题意:给出一个数,问是否有ai + bj + ck == x 二分查找:首先计算sum[l] = a[i] + b[j],对于q,枚举ck,查找是否有sum + ck == x */ #include <cstdio> #include <algorithm> #include <cmath> using namespace std; typedef long long ll; ; const int INF = 0x3f3f3f3f; ll a[MA…

Vases and Flowers 题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=4614 Problem Description Alice is so popular that she can receive many flowers everyday. She has N vases numbered from 0 to N-1. When she receive some flowers, she will try to put them…

Can you find it? Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)Total Submission(s): 17036    Accepted Submission(s): 4337 Problem Description Give you three sequences of numbers A, B, C, then we give you a number…

OO's Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 955    Accepted Submission(s): 358 Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the nu…

Doubles Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4266    Accepted Submission(s): 2945 Problem Description As part of an arithmetic competency program, your students will be given randoml…

题意:给出一个整数nnn, 找出一个大于等于nnn的最小整数mmm, 使得mmm可以表示为2a3b5c7d2^a3^b5^c7^d2​a​​3​b​​5​c​​7​d​​. 析:预处理出所有形为2a3b5c7d2^a3^b5^c7^d2​a​​3​b​​5​c​​7​d​​即可, 大概只有5000左右个.然后用二分查找就好. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio…

题意: 给定一个素数p(p <= 1e12),问是否存在一对立方差等于p. 分析: 根据平方差公式: 因为p是一个素数, 所以只能拆分成 1*p, 所以 a-b = 1. 然后代入a = b + 1. 求出 3a² + 3a + 1 = p 化简得a(a+1) = (p-1)/3 令(p-1)/3 = T, 问题化为是否存在整数a使得a(a+1) == T, 那么令 t = (int)sqrt(T),只要判定一下t * (t+1) == T ? 即可 另一种做法是打一个a的表(a只要打到1e6)…

解题思路:给出两个数列an,bn,求an和bn中相同元素的个数因为注意到n的取值是0到1000000,所以可以用二分查找来做,因为题目中给出的an,bn,已经是单调递增的,所以不用排序了,对于输入的每一个b[i],查找它在an数列中是否存在.存在返回值为1,不存在返回值为0 CD Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 627…

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1597 find the nth digit Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9517    Accepted Submission(s): 2757 Problem Description 假设:S1 = 1S2 = 12S3…

http://acm.hdu.edu.cn/showproblem.php?pid=4614 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Problem Description Alice is so popular that she can receive many flowers everyday. She has N vases numbered from 0 to…

第一种:顺序查找法 中心思想:和数组中的值逐个比对! /* * 参数说明: * array:传入数组 * findVal:传入需要查找的数 */ function Orderseach(array,findVal){ var temp = false; //控制开关 for(var i =0;i<array.length;i++){ if(array[i] == findVal){ //逐个匹配是否相等 temp = true; //如果找到,temp设置为true; return i; //返…

二分查找又称折半查找,它是一种效率较高的查找方法. 折半查找的算法思想是将数列按有序化(递增或递减)排列,查找过程中采用跳跃式方式查找,即先以有序数列的中点位置为比较对象,如果要找的元素值小 于该中点元素,则将待查序列缩小为左半部分,否则为右半部分.通过一次比较,将查找区间缩小一半. 折半查找是一种高效的查找方法.它可以明显减少比较次数,提高查找效率.但是,折半查找的先决条件是查找表中的数据元素必须有序. 折半查找法的优点是比较次数少,查找速度快,平均性能好;其缺点是要求待查表为有序表,且插入删…

二分法的基本思路是对一个有序序列(递增递减都可以)查找时,测试一个中间下标处的值,若值比期待值小,则在更大的一侧进行查找(反之亦然),查找时再次二分.这比顺序访问要少很多访问量,效率很高. 设:low,hight,mid均为整型.以在一个降序arr[5]={5,4,2,1,0}中查找k=4时的下标为例,取low=0,hight=4,则mid=low+(hight-low)/2=2(若无溢出可直接相加取半),此时arr[mid]=2小于k,这时需要向值更大的一侧(左侧)查找,所以low不变,hig…

/** * 二分查找 * @param a * @param n * @param value * @return * @date 2016-10-8 * @author shaobn */ public static int binaryFind(int[] a,int n,int value){ int lowNum = 0; int highNum = n-1; while(lowNum<=highNum){ int midNum = (lowNum+highNum)/2; if(a[mi…

最新IP地址数据库  来自 qqzeng.com 利用二分逼近法(bisection method) ,每秒300多万, 比较高效! 原来的顺序查找算法 效率比较低 readonly string ipBinaryFilePath = "qqzengipdb.dat"; readonly byte[] dataBuffer, indexBuffer; ]; readonly int dataLength; public IpLocation() { try { FileInfo fil…

折半搜索,也称二分查找算法.二分搜索,是一种在有序数组中查找某一特定元素的搜索算法. A 搜素过程从数组的中间元素开始,如果中间元素正好是要查找的元素,则搜素过程结束: B 如果某一特定元素大于或者小于中间元素,则在数组大于或小于中间元素的那一半中查找,而且跟开始一样从中间元素开始比较. C 如果在某一步骤数组为空,则代表找不到.这种搜索算法每一次比较都使搜索范围缩小一半. 时间复杂度折半搜索每次把搜索区域减少一半,时间复杂度为. (n代表集合中元素的个数)空间复杂度 /// <summary>…

二分查找:在一段数字内,找到中间值,判断要找的值和中间值大小的比较.如果中间值大一些,则在中间值的左侧区域继续按照上述方式查找.如果中间值小一些,则在中间值的右侧区域继续按照上述方式查找.直到找到我们希望的数字. def search_data(data,data_find): # 中间值的索引号的定义:数组长度/2 mid = int(len(data)/2) # 判断从1开始的数字数组内查找 if data[mid] >= 1: # 如果我们要找的值(data_find)比中间值(data[…

PHP实现文本快速查找 - 二分查找法 起因 先说说事情的起因,最近在分析数据时经常遇到一种场景,代码需要频繁的读某一张数据库的表,比如根据地区ID获取地区名称.根据网站分类ID获取分类名称.根据关键词ID获取关键词等.虽然以上需求都可以在原始建表时,通过冗余数据来解决.但仍有部分业务存的只是关联表的ID,数据分析时需要频繁的查表. 所读的表存在共同的特点 数据几乎不会变更 数据量适中,从一万到100多万,如果全加载到内存也不太合适. 纠结的地方 在做数据分析时,需要十分频繁的读这些表,每秒有可…

最近做笔试题有这么一个关于二分查找的例子. 给一个有序数组,和一个查找目标,用二分查找找出目标所在index,如果不存在,则返回-1-(其应该出现的位置),比如在0,6,9,15,18中找15,返回3:找10.则返回-4(-1-3) 实现如下: public class Sulution1 { public static void main(String[] args) { System.out.println(findBySep(2, new int[]{0,2,4,6,9})); } pub…

给数组赋值:通过fill方法. 对数组排序:通过sort方法,按升序.比较数组:通过equals方法比较数组中元素值是否相等.查找数组元素:通过binarySearch方法能对排序好的数组进行二分查找法操作. 使用如下: int[] array = new int[5]; //填充数组 Arrays.fill(array, 5); System.out.println("填充数组:Arrays.fill(array, 5):"); test.output(array); //将数组的第…

约12年年底的时候,接触了python不到半年的样子,入门是直接实现GUI测试case的.今天面试地平线机器人,发现忘得差不多了- -. 当时的问题是这样的 写一个二分查找是实现,我好像不记得二分查找是个啥- -面试官很nice的解释了一遍.当时的写法是这样的. #!/bin/usr/env python inputArr=[1,2,3,4,4,5,5,5,6,6,6,7,77,77] destStr=sys.argv[1] if inputArr.find(destStr) == -1: pr…