UVA.136 Ugly Numbers (优先队列) 题意分析 如果一个数字是2,3,5的倍数,那么他就叫做丑数,规定1也是丑数,现在求解第1500个丑数是多少. 既然某数字2,3,5倍均是丑数,且1为丑数,那么不妨从1开始算起.算完之后2,3,5均为丑数,然后再从2算起,4,5,10均为丑数--直到算到第1500个即可.那么有如下的问题: 如何从算出来的丑数中取出最小的? 如何保证没有计算重复的丑数? 对于第一个问题,可以采用优先队列的方法,即有序的队列,且为升序,每次只需要取出队首元素即可…

Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... shows the first 11 ugly numbers. By convention, 1 is included. Write a program to find and print the 1500’th ugly number. Input Ther…

Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... shows the first 11 ugly numbers. By convention, 1 is included.Write a program to find and print the 1500'th ugly number. Input There…

题意:给出丑数的定义,不能被除2,3,5以外的素数整除的的数称为丑数. 和杭电的那一题丑数一样--这里学的紫书上的用优先队列来做. 用已知的丑数去生成新的丑数,利用优先队列的能够每次取出当前最小的丑数再去生成新的丑数==== 大概这儿的优先队列就充当了dp转移方程里面的那个min的意思@_@ #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include&l…

题目大意:只有素因子2,3,5的数叫做丑数.输出第1500个丑数即可. 这个...好吧,直接输出就是了.自己写一个小程序先计算一下,这就是黑盒测试的好处啊,“我们的目标是解决问题,而不是为了写程序而写程序,同时应该保持简单(Kepp It Simple and Stupid, KISS)”,摘自<算法竞赛入门经典>. #include <cstdio> int main() { ; // the number of ugly number int i; ; ; i++) { int…

 Ugly Numbers  Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... shows the first 11 ugly numbers. By convention, 1 is included. Write a program to find and print the 1500'th ugly num…

题目链接: https://cn.vjudge.net/problem/UVA-136 /*问题 输出第1500个丑数,丑数的定义是不能被2,3,5以外的其他素数整除的数 解题思路 直接硬暴力先试一下发现枚举的时候不知道它能被那个素数整除,况且打很大的素数表依次遍历也不一定能够找到1500个丑数. 那么根据定义,如果x是一个丑数,那么2*x.3*x.5*x也一定是丑数.那么构建一个优先队列,让数值小的优先级最高,加入第一个丑数1, 依次加入三个丑数(数值小的在前面),直到第1500个丑数时结束,…

Problem Description Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ...shows the first 11 ugly numbers. By convention, 1 is included. Write a program to find and print the 1500’th ugly…

丑数(Ugly Numbers, UVa 136) 题目描述 我们把只包含因子2.3和5的数称作丑数(Ugly Number).求按从小到大的顺序的第1500个丑数.例如6.8都是丑数,但14不是,因为它包含因子7.习惯上我们把1当做第一个丑数. 算法实现 版本1:错误版本 //#define LOCAL #include<iostream> #include<cstdio> #include<queue> /* 输出第1500个丑数 */ using namespac…

Ugly Numbers Descriptions: Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... shows the first 11 ugly numbers. By convention, 1 is included. Write a program to find and print the 1500…

Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... shows the first 11 ugly numbers. By convention, 1 is included. Write a program to find and print the 1500'th ugly number. Input Ther…

Ugly Numbers 题目 Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... shows the first 11 ugly numbers. By convention, 1 is included. Write a program to find and print the 1500'th ugly nu…

题目链接:http://poj.org/problem?id=1338 Ugly Numbers Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 21920   Accepted: 9789 Description Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10,…

Ugly Numbers Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 21918 Accepted: 9788 Description Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, - shows the first 10 ugly numbers.…

Ugly Numbers 时间限制:1000 ms  |  内存限制:65535 KB 难度:2 描写叙述 Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... shows the first 11 ugly numbers. By convention, 1 is included. Now give you ne…

[POJ1338]Ugly Numbers 试题描述 Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, ... shows the first 10 ugly numbers. By convention, 1 is included. Given the integer n,write a program to find an…

UVa 10006 - Carmichael Numbers An important topic nowadays in computer science is cryptography. Some people even think that cryptography is the only important field in computer science, and that life would not matter at all without cryptography. Alvar…

转载请注明出处:viewmode=contents">http://blog.csdn.net/u012860063? viewmode=contents 题目链接:id=1338">http://poj.org/problem?id=1338 Description Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence  1, 2, 3, 4, 5, 6, 8, 9, 1…

https://leetcode.com/problems/ugly-number/ Write a program to check whether a given number is an ugly number. Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 6, 8 are ugly while 14 is not ugly since it include…

Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, …shows the first 11 ugly numbers. By convention, 1 is included.Write a program to find and print the 150′th ugly number. METHOD 1 (Simple…

题目链接:uva 12050 - Palindrome Numbers 题意:求第n个回文串 思路:首先可以知道的是长度为k的回文串个数有9*10^(k-1),那么依次计算,得出n是长度为多少的串,然后就得到是长度为多少的第几个的回文串了,有个细节注意的是, n计算完后要-1! 下面给出AC代码: #include <bits/stdc++.h> typedef long long ll; using namespace std; ; ll num[maxn]; int n,ans[maxn]…

一.题目大意 本题要求写出前1500个仅能被2,3,5整除的数. 二.题解 最初的想法是从1开始检验该数是否只能被2,3,5整除,方法是这样的,对于一个数,如果它能被2整除,就除以2,如果它能被3整除,就除以3,如果它能被5整除,就除以5,直到不能被2,3,5整除,看结果是不是1,如果是1就满足条件,否则不满足条件.但是第1500个数大约近10亿,显然是1s内不可以完成的. 然后仔细分析发现:满足条件的数是2^x*3^y*5^z,可以转换为这样的定义,定义一个集合,该集合中有元素1,如果x在集合…

UVA - 13022 Sheldon Numbers 二进制形式满足ABA,ABAB数的个数(A为一定长度的1,B为一定长度的0). 其实就是寻找在二进制中满足所有的1串具有相同的长度,所有的0串也具有相同的长度,并且在给定范围内的个数. 位运算.通过分析不难发现,所有解不会很大,因此我们可以暴力,用两个for分别枚举0串和1串的长度,然后交替放入一个值内(注意先放1),同时更新答案. 将值存入set,发现所有满足条件的个数为4809(4810). ps:因为这里的2^63大于long lon…

题目: Ugly numbers are numbers whose only prime factors are 2, 3 or 5 . The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... shows the first 11 ugly numbers. By convention, 1 is included. Write a program to find and print the 1500’th ugly number. Input…

A. Bad Ugly Numbers time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given a integer nn (n>0n>0). Find any integer ss which satisfies these conditions, or report that there are no s…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 每个丑数x,都能生成3个丑数2x,3x,5x 则我们以1作为起点. 生成丑数. 每次取出set里面最小的那个数. 然后用它去生成其他的丑数. 加入到set里面. 然后再把set的头元素删掉就好. 取1500次. 不知道生成的哪一个更大,那么每次取最小的就好了,set就是用来做这个事的 每次都取最小的来生成,那么新生成的数字,肯定也是尽可能地小的,所以不会出现取出来的不是第i小的数的情况 [代码] #include <bits/s…

丑数 丑数是指不能被2,3,5以外的其他素数整除的数.把丑数从小到大排列起来,结果如下: 1,2,3,4,5,6,8,9,10,12,15,… 求第1500个丑数. 提示:从小到大生成各个丑数.最小的丑数是1,对于任意丑数x,2x,3x和5x也是丑数.使用一个优先队列保存已生成的丑数,每次取出最小的丑数,生成3个新的丑数.需要注意,同一个丑数有多种生成方式,所以需要判断一个丑数是否已经生成过. 题目链接:https://vjudge.net/contest/211547#problem/G #i…

题目链接:uva 1422 - Processor 题目大意:有一个机器要处理一些问题,给出这些问题可以开始的时间和必须完成的时间,以及任务的工作量,问说机器必须以最少每秒多少得工作量才能完成这些任务. 解题思路:二分答案,答案的上限是MaxWork * MaxN(全在一秒内完成).然后就是判断说在c的工作量小是否能完成任务.因为处理一个问题的时间可以不连续,所以给解题增加了难度. 现将任务按照开始的时间从小到大排序,然后枚举时间区间.开始时间比当前枚举时间小的话入队. 队列用优先队列,结束时间…

题意  求把全部数加起来的最小代价  a+b的代价为(a+b) 越先运算的数  要被加的次数越多  所以每次相加的两个数都应该是剩下序列中最小的数  然后结果要放到序列中  也就是优先队列了 #include<cstdio> #include<queue> using namespace std; priority_queue<int, vector<int>, greater<int> >q; typedef long long ll; ll…

优先队列. 做法:维护两个优先队列:quesell  和  quebuy, 一个是小值优先,一个是大值优先.每次push的时候,都取各自的Top元素,比较价格,如果卖的比卖的出价低,则成交,各自的要买和要卖的股票数量减少能够减少的最大值,此时的DP(DealPrice)被记录下来. 具体见代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include…