称号:发现没有比给定数量少n的.只要2,3一个因素的数字组成. 分析:数论.贪婪,分而治之. 用两个三分球,分别代表乘法2,和繁殖3队列,队列产生的数字,原来{1}. 然后.每取两个指针相应元素*2和*3的值中最小的即为未找到的数字中最小的: 注意,可能生成反复数据.不要存进去(反复数据.一定连续产生). 说明:打表计算.二分查询输出就可以. #include <iostream> #include <cstdlib> #include <cstdio> using n…

 Krypton Factor  You have been employed by the organisers of a Super Krypton Factor Contest in which contestants have very high mental and physical abilities. In one section of the contest the contestants are tested on their ability to recall a seque…

题目:输出n!中素数因数的个数. 分析:数论.这里使用欧拉筛法计算素数,在计算过程中求解就可以. 传统筛法是利用每一个素数,筛掉自己的整数倍: 欧拉筛法是利用当前计算出的全部素数,乘以当前数字筛数: 所以每一个前驱的素椅子个数一定比当前数的素因子个数少一个. 说明:重新用了"线性筛法". #include <algorithm> #include <iostream> #include <cstdlib> #include <cstring&g…

Time limit 3000 ms OS Linux Write a program, that computes the number of different prime factors in a positive integer.InputThe input tests will consist of a series of positive integers. Each number is on a line on its own. Themaximum value is 100000…

 Factors and Factorials  The factorial of a number N (written N!) is defined as the product of all the integers from 1 to N. It is often defined recursively as follows: Factorials grow very rapidly--5! = 120, 10! = 3,628,800. One way of specifying su…

https://vjudge.net/problem/UVA-1575 题意: 令f(k)=n 表示 有n种方式,可以把正整数k表示成几个数的乘积的形式. 例 10=2*5=5*2,所以f(10)=2 给出n,求最小的k 搜索 从最小的质数开始枚举选几个 假设前i-1个种质数用了k个,有sum种方案,第i种质数选a个, 那么前i种质数的方案就有sum*C[k+a][a] 可以理解原来有k个位置,又加了a个位置,有a个数可以放在任意位置 所以前i种的每一种方案都变成C[k+a][a]种 枚举每个质…

题目大意:Euler's Totient的应用. 几乎和UVa 10179 - Irreducable Basic Fractions一样,于是偷了个懒,直接用10179题的代码,结果WA了,感觉一样啊...然后就搜,看到n=1的特殊情况,好吧,读题的时候也小小注意了一下"less than" 呢,写代码时就忘的一干二净了...读题时应该把注意事项记下来的... #include <cstdio> #include <vector> #include <a…

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…

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UVA - 10564 Paths through the Hourglass 题意: 要求从第一层走到最下面一层,只能往左下或右下走 问有多少条路径之和刚好等于S? 如果有的话,输出字典序最小的路径. f[i][j][k]从下往上到第i层第j个和为k的方案数 上下转移不一样,分开处理 没必要判断走出沙漏 打印方案倒着找下去行了,尽量往左走   沙茶的忘注释掉文件WA好多次   #include <iostream> #include <cstdio> #include <a…