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Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n). If d(a) = b and d(b) = a, where a b, then a and b are an amicable pair and each of a and b are called amicable numbers. For example, the proper d…
Problem 21 https://projecteuler.net/problem=21 Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n). If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are c…
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 # Definition for singly-link…
题目: 小O是一个热爱短代码的选手.在缩代码方面,他是一位身经百战的老手.世界各地的OJ上,很多题的最短解答排行榜都有他的身影.这令他感到十分愉悦. 最近,他突然发现,很多时候自己的程序明明看起来比别人的更短,实际代码量却更长.这令他感到很费解.经过一番研究,原来是因为他每一行的缩进都全是由空格组成的,大量的空格让代码量随之增大. 现在设小O有一份 \(n\) 行的代码,第 \(i\) 行有 \(a_i\) 个空格作为缩进. 为解决这一问题,小O要给自己文本编辑器设定一个正整数的默认TAB宽度…
题链: http://uoj.ac/problem/21 题解: ...技巧题吧 先看看题目让求什么: 令$F(x)=\sum_{i=1}^{n}(\lfloor a[i]/x \rfloor +a[i]$%$x)$ 要求输出最小的F(x). 首先不难看出,x的取值不会超过最大的a[i]+1,(因为之后的答案都和x==a[i]+1时的答案相同) 把式子化为如下形式: $F(x)=\sum_{i=1}^{n}(\lfloor a[i]/x \rfloor +(a[i]-\lfloor a[i]/x…
传送门 http://uoj.ac/problem/21 枚举 (调和级数?) $\sum_{i=1}^{n} (a_i / x + a_i \bmod x) =\sum a_i - (\sum_{i=1}^{n} a_i /x) * (x-1)$ 看上去并没有一个很好的办法确定x的取值? 大概只能暴力枚举了. 枚举x的大小,如果用分块加速的方法统计解,复杂度是O(n)+O(n/2)+O(n/3)+O(n/4)+... 累积起来是O(nlogn) 嗯?好像是正解? イミワカナイ #include…
There are some Common Bugs in C Programming. Most of the contents are directly from or modified from Prof. Liu Pangfeng's blog. Most credits should go to him. For all the following problems, answer the output message of the code, unless they are spec…
关于 caesum.com 网上上的题目,分类有Sokoban,Ciphers,Maths,Executables,Programming,Steganography,Misc.题目有点难度,在努力奋战中…… problem 21 Factor 数806515533049393最大的质因数是多少? 因数分解题,32位int范围在四十亿左右.我是这么记这个大概值的,全球六十多亿人口,很多人还不能上网,而且有局域网的存在,IP本来够用的,因为北美划用了很多,导致了现在的IP不够用,出现了IPv6.6…
A. Harmonic Matrix Counter (3/19) B. Binary Tree (1/14) C. Asa's Chess Problem (21/65) [ Problem ] 给出一个棋盘格,里面的格子两两分为一组,组内的格子可以交换位置, 同一组格子一定在同一行或者同一列,每个格子为白色或者黑色, 现在要求每行中黑色的格子数量在rL到rR之间,每列中黑色的格子数量在cL到cR之间, 现在要求求出最少的交换次数使得要求被满足,不能满足则输出-1 [ Solution ] 我…
http://oj.xjtuacm.com/problem/21/ 对13进行分析,每种价格出现的次数: $(C_m^1 + C_m^2 + ... + C_m^m)(C_{n - m}^0 + C_{n - m}^1 + C_{n - m}^2 + ... + C_{n - m}^{n - m}) = ({2^m} - 1)*{2^{n - m}} = {2^n} - {2^m}$ #include<cstdio> #include<cstring> #include<cs…
UOJ_21_[UR #1]缩进优化_数学 题面:http://uoj.ac/problem/21 最小化$\sum\limits{i=1}^{n}a[i]/x+a[i]\;mod\;x$ =$\sum\limits{i=1}^{n}(1-x)*(a[i]/x)+a[i]$ =$\sum\limits{i=1}^{n}a[i]-\sum\limits{i=1}^{n}(x-1)*a[i]/x$ 直接枚举x,然后对于0~x-1这部分贡献是0,x~2x-1这部分贡献是1. 按x分块计算即可. 代码:…
Levmar:Levenberg-Marquardt非线性最小二乘算法 eryar@163.com Abstract. Levmar is GPL native ANSI C implementations of the Levenberg-Marquardt optimization algorithm.The blog focus on the compilation of levmar on Windows with Visual Studio. Key Words. Levmar, C,…
All LeetCode Questions List 题目汇总 Sorted by frequency of problems that appear in real interviews. Last updated: October 2, 2017Google (214)534 Design TinyURL388 Longest Absolute File Path683 K Empty Slots340 Longest Substring with At Most K Distinct C…
本题来自 Project Euler 第21题:https://projecteuler.net/problem=21 ''' Project Euler: Problem 21: Amicable numbers Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n). If d(a) = b and d(b) = a, where a ≠ b…
代码: %% ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ %% Output Info about this m-file fprintf('\n***********************************************************\n'); fprintf(' <DSP using MATLAB> Problem 6.21 \n\n'); ba…
证明: 代码: %% ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ %% Output Info about this m-file fprintf('\n***********************************************************\n'); fprintf(' <DSP using MATLAB> Problem 5.2…
快到龙抬头,居然下雪了,天空飘起了雪花,温度下降了近20°. 代码: %% ------------------------------------------------------------------------ %% Output Info about this m-file fprintf('\n***********************************************************\n'); fprintf(' <DSP using MATLAB>…
代码: %% ------------------------------------------------------------------------ %% Output Info about this m-file fprintf('\n***********************************************************\n'); fprintf(' <DSP using MATLAB> Problem 8.21 \n\n'); banner();…
H - Solve this interesting problem Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 5323 Appoint description:  System Crawler  (2015-07-28) Description Have you learned something about segment tr…
[263]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 includes another prime factor 7.…
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Vasya has an array a consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the seco…
Problem A Lucky Year 题目传送门[here] 题目大意是说,只有一个数字非零的数是幸运的,给出一个数,求下一个幸运的数是多少. 这个幸运的数不是最高位的数字都是零,于是只跟最高位有关,只保留最高位,然后加一求差就是答案. Code /** * Codeforces * Problem#808A * Accepted * Time:15ms * Memory:0k */ #include<iostream> #include<cstdio> #include<…
The 3n + 1 problem Problem Description Problems in Computer Science are often classified as belonging to a certain class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose classificat…
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Digital collectible card games have become very popular recently. So Vova decided to try one of these. Vova has n cards in his collection. Each of these cards is characterised by its power pi, magic number ci and level li. Vova wants to build a deck…
模拟信号经过不同的采样率进行采样后,得到不同的数字角频率,如下: 三种Fs,采样后的信号的谱 重建模拟信号,这里只显示由第1种Fs=0.01采样后序列进行重建,采用zoh.foh和spline三种方法:…
题意:给你一棵树,每个点有一个左括号或者右括号,问你树上能够完美匹配的路径数量(l->r,r->l 视作不同路径). 点分治可以使用“不扣去重复答案”的写法,只不过,要先将每个点的子树按照从小到大的顺序排序,防止复杂度出错.(此题不需要,因为统计一个子树的贡献的时候,时间复杂度最多只与当前子树的大小有关,与之前的无关) 要将重心归到“之前的所有子树中”,而“当前的子树”不含重心. 把左括号视作1,右括号视作-1. 对于每颗子树中的点,统计三个数组,分别表示重心下面的结点到当前结点的路径的前缀和…
#include<stdio.h> int main() { float f,c; while(scanf("%f",&f)!=EOF) c=*(f-); printf("%.3f",c); ; }…
#include<stdio.h> int main() { int a,b,c; scanf("%d %d %d",&a,&b,&c); printf("%.3f\n",(a+b+c)/3.0); ; }…