145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145. Find the sum of all numbers which are equal to the sum of the factorial of their digits. Note: as 1! = 1 and 2! = 2 are not sums they are not included. 题目大意: 145 是一个奇怪的数字, 因为 1! + 4! + 5!…

Problem 34 https://projecteuler.net/problem=34 145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145. 145是一个神奇的数字,1! + 4! + 5! = 1 + 24 + 120 = 145. Find the sum of all numbers which are equal to the sum of the factorial of their digits. 找到所有…

145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145. Find the sum of all numbers which are equal to the sum of the factorial of their digits. Note: as 1! = 1 and 2! = 2 are not sums they are not included. #include <iostream> #include <v…

Problem 34 145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145. Find the sum of all numbers which are equal to the sum of the factorial of their digits. Note: as 1! = 1 and 2! = 2 are not sums they are not included. puts (0..50000).select{|…

题链: http://uoj.ac/problem/34 题解: FFT入门题. (终于接触到迷一样的FFT了) 初学者在对复数和单位根有简单了解的基础上,可以直接看<再探快速傅里叶变换>(毛啸). (主要用于求两个序列的卷积) 代码: 递归版: #include<bits/stdc++.h> #define MAXN 300000 using namespace std; const double Pi=acos(-1); struct Z{ double real,image;…

http://uoj.ac/problem/34 看了好长时间的FFT和NTT啊qwq在原根那块磨蹭了好久_(:з」∠)_ 首先设答案多项式的长度拓展到2的幂次后为n,我们只要求出一个g(不是原根)满足\(i\in \{1\dots n\},g^i\)互不相同,且\(g^n=1\). 把这个g当做"FFT里面的主n次单位根"的类似物. 而且\(g^{\frac n2}=-1\),因为\(g^{\frac n2}\)与\(g^n\)不相同且\((g^{\frac n2})^2=g^n=1…

学习内容:国家集训队2016论文 - 再谈快速傅里叶变换 模板题:http://uoj.ac/problem/34 1.基本介绍 对长度为L的\(A(x),B(x)\)进行DFT,可以利用 \[ \begin{align} P(x)=A(x)+iB(x) \tag{1} \\ Q(x)=A(x)-iB(x) \tag{2} \end{align} \] 对\(P(x)\)进行DFT,得到\(F_p\). \(Q(x)\)的结果 DFT\(F_q[k]=!(F_p[2L-k])\),(!表示取共轭…

http://uoj.ac/problem/34 (题目链接) 题意 求两个多项式的乘积 Solution 挂个FFT板子. 细节 FFT因为要满足$n$是$2$的幂,所以注意数组大小. 代码 // uoj34 #include<algorithm> #include<iostream> #include<cstdlib> #include<cstring> #include<complex> #include<cstdio> #in…

本文版权归ljh2000和博客园共有,欢迎转载,但须保留此声明,并给出原文链接,谢谢合作. 本文作者:ljh2000 作者博客:http://www.cnblogs.com/ljh2000-jump/转载请注明出处,侵权必究,保留最终解释权! 题目链接:http://uoj.ac/problem/34 正解:FFT 解题报告: 非递归版FFT模板保存. //It is made by ljh2000 #include <iostream> #include <cstdlib> #i…

摘要 多任务学习(Multi-Task Learning, MTL)是机器学习中的一种学习范式,其目的是利用包含在多个相关任务中的有用信息来帮助提高所有任务的泛化性能. 首先,我们将不同的MTL算法分为特征学习法.低秩方法.任务聚类方法.任务关系学习方法和分解方法,然后讨论每种方法的特点.为了进一步提高学习任务的性能,MTL可以与其他学习范式相结合,包括半监督学习.主动学习.非监督学习.强化学习.多视图学习和图形模型.当任务数量较大或数据维数较高时,批量MTL模型难以处理,本文对在线.并行和分布…