It is a small fun problem to solve. Since only a max sum is required (no need to print path), we can only keep track of the max value at each line. Basically it is still a human labor simulation work. To be more specifically, we need keep track of a…

Let's consider a triangle of numbers in which a number appears in the first line, two numbers appear in the second line, three in the third line, etc. Develop a program which will compute the largest of the sums of numbers that appear on the paths st…

传送门 这次fftfftfft乱搞居然没有被卡常? 题目简述:给你nnn个数,每三个数ai,aj,ak(i<j<k)a_i,a_j,a_k(i<j<k)ai​,aj​,ak​(i<j<k)组成的所有和以及这些和出现的次数. 读完题直接让我联想到了昨天写过的一道用fftfftfft优化点分治合并的题 ,这不是差不多嘛? 只是这一次的容斥要麻烦一些. 我们令原数列转化成的系数序列为{an}\{a_n\}{an​} 那么如果允许重复答案就应该是an3a_n^3an3​ 然后展…

# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <complex> # include <set> # includ…

[传送门] FFT第一题! 构造多项式 $A(x) = \sum x ^ {s_i}$. 不考虑题目中 $i < j < k$ 的条件,那么 $A^3(x)$ 每一项对应的系数就是答案了. 考虑容斥. $$(\sum x)^3 = \sum x^3 + 3 \sum x^2 y + 6\sum xyz$$ $$(\sum x^2) (\sum x)= \sum x^3 + \sum x^2 y$$ 所以 $$\sum xyz = \dfrac{(\sum x)^3 - 3 (\sum x^2)…

本题有两个难点: 1 大量的数据输入.没处理好就超时 - 这里使用buffer解决 2 因子分解的算法 a)暴力法超时 b)使用sieve(筛子),只是当中的算法逻辑也挺不easy搞对的. 数值N因子分解逻辑: 1 保存全部能够sqrt(N)范围内的质素 2 找到能够被N除尽的质素d, 然后用d去除N.使用deg变量,保存度.即有多少个d能够被N除尽 3 用d去乘全部已经找到的因子(包含1),假设度deg大于1.那么循环i从1到deg, 用d*i去乘全部找到的因子 找到全部因子相加,减去N,就是…

文章目录 多项式的运算 多项式的加减法,数乘 多项式乘法 多项式求逆 多项式求导 多项式积分 多项式取对 多项式取exp 多项式开方 多项式的除法/取模 分治FFT 生成函数 相关题目 多项式的运算 多项式的加减法,数乘 这个大家应该都会吧 不推了. 直接上代码: friend inline poly operator+(const poly&a,const poly&b){ poly ret(max(a.deg(),b.deg())); for(ri i=0;i<=a.deg();…

题目 Source http://www.spoj.com/problems/TSUM/ Description You're given a sequence s of N distinct integers.Consider all the possible sums of three integers from the sequence at three different indicies.For each obtainable sum output the number of diff…

You're given a sequence s of N distinct integers.Consider all the possible sums of three integers from the sequence at three different indicies.For each obtainable sum output the number of different triples of indicies that generate it.Constraints:N…

题目链接 首先忽略 i < j < k这个条件.那么我们构造多项式$$A(x) = \sum_{1现在我们考虑容斥:1. $ (\sum_{}x)^3 = \sum_{}x^3 + 3\sum_{}x^2 y + 6\sum_{}xyz $ 2. $ (\sum_{}x^2)(\sum_{}x) = \sum_{}x^3 + \sum_{}x^2 y $3. $ (\sum_{}x)^3 = \sum_{}x^3 $由上面三个式子 我们可以推导出$ \sum_{}xyz = \frac {(\…

题意:给你n个数字,对于任意s,s满足\(s=u_i+u_j+u_k,i<j<k\),要求出所有的s和对应满足条件的i,j,k的方案数 Solution: 构造一个函数:\(A(x)=\sum_{i=0}^{n-1}a_ix^i\),这是一个多项式 对于每一个\(u_i\),我们把这个多项式中的\(x^{u_i}\)的系数\(a_{u_i}\)加上一 也就是说,对于任意\(x^i\),它的系数为i在给出序列中出现的次数 多项式的三次方为: \[ C(x)=A(x)^3\\ C(x)=\sum_…

原地址http://www.starlino.com/dcm_tutorial.html Introduction This article is a continuation of my IMU Guide, covering additional orientation kinematics topics. I will go through some theory first and then I will present a practical example with code bui…

http://www.spoj.com/problems/SEGSQRSS/ SPOJ Problem Set (classical) 11840. Sum of Squares with Segment Tree Problem code: SEGSQRSS Segment trees are extremely useful.  In particular "Lazy Propagation" (i.e. see here, for example) allows one to c…

Triangular Sums 时间限制:3000 ms  |  内存限制:65535 KB 难度:2   描述 The nth Triangular number, T(n) = 1 + - + n, is the sum of the first n integers. It is the number of points in a triangular array with n points on side. For example T(4): X X X X X X X X X X Wr…

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below. For example, given the following triangle [ [2], [3,4], [6,5,7], [4,1,8,3] ] The minimum path sum from top to bottom is 11 (i…

ZC:本来以为没有中文版的,原来有中文版,网址为: ZC: OpenGL3.0教程 _ 泰然网.html(http://www.tairan.com/archives/6126/) ZC: OpenGL3.0教程 第三课: 矩阵 _ 泰然网.html(http://www.tairan.com/archives/5951/) 1.网址: Basic OpenGL.html (http://www.opengl-tutorial.org/beginners-tutorials/) Tutorial…

MSDN中的一篇博文链接:Tutorial series: learning how to write a 3D soft engine from scratch in C#, TypeScript or JavaScript This tutorial is part of the following series: 1 – Writing the core logic for camera, mesh & device object (this article) 2 – Drawing li…

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2640 http://www.spoj.com/problems/SPOINTS/en/ http://poj.org/problem?id=3805 http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1298 要…

Triangular Sums 时间限制:3000 ms  |  内存限制:65535 KB 难度:2   描述 The nth Triangular number, T(n) = 1 + … + n, is the sum of the first n integers. It is the number of points in a triangular array with n points on side. For example T(4): XX X X X X X X X X Wri…

Make Triangle Chayanika loves Mathematics. She is learning a new chapter geometry. While reading the chapter a question came in her mind. Given a convex polygon of n sides. In how many ways she can break it into triangles, by cutting it with (n-3) no…

122-Triangular Sums 内存限制:64MB 时间限制:3000ms 特判: No 通过数:5 提交数:7 难度:2 题目描述: The nth Triangular number, T(n) = 1 + … + n, is the sum of the first n integers. It is the number of points in a triangular array with n points on side. For example T(4): X X X X…

FZU Monthly-201905 tutorial 题目(难度递增) easy easy-medium medium medium-hard hard 思维难度 AB H DG CE F A. Chess 一道模拟题,很多人做不出来的原因是没有看懂题意. 题意重述: 商店中给出了一个LV1棋子出现的顺序序列,我们需要按照这个序列来购买所有的棋子. 我们最多可以存下16个棋子,每个棋子有一个种类,当有3个相同的棋子的时候,棋子就会升级,3个LV1变成LV2.3个LV2变成LV3. 当一种棋到达…

今天一如既往地在luogu刷题,发现了一个新OJ,就去网上查资料.得到了下面这些. 以下转载自这里 SPOJ是波兰最为出色的Online Judge之一,界面和谐,题目类型也非常丰富,适合有一定基础的选手练习,对高手而言也是个提高能力的良好平台. SPOJ题目分类:classical,challenge,partial,tutorial. 1)classical:ACM题型,通过所有数据才能算AC 2)challenge:有趣的题目,每个题目有不同的评分标准(代码长短,效果好坏,速度等),感觉都…

按tutorial打的我血崩,死活挂第四组- - 思路来自FXXL /* CodeForces 837F - Prefix Sums [ 二分,组合数 ] | Educational Codeforces Round 26 题意: 设定数组 y = f(x) 使得 y[i] = sum(x[j]) (0 <= j < i) 求初始数组 A0 经过多少次 f(x) 后 会有一个元素 大于 k 分析: 考虑 A0 = {1, 0, 0, 0} A1 = {1, 1, 1, 1} -> {C(…

和泉纱雾与烟花大会 题目来源: UOJ 192 最强跳蚤 (只改了数据范围) 官方题解: 在这里哦~(说的很详细了 我都没啥好说的了) 题目大意: 求树上各边权乘积是完全平方数的路径数量. 这种从\(n^2\)条路径中找出满足xx条件的路径的条数的题, 我们可以根据常识判断要用到点分治. 不过这题并没有用到点分治, 这个一会再说, 我们先来看部分分. 哎呀其实这题好多部分分我都不会写(捂脸 算法1: 直接乘边权处理显然是不行哒, 怕是\(w\leq2\)怕是都要用到高精度了(什么你说\(w\le…

Floyd's Triangle Floyd's triangle is a right-angled triangular array of natural numbers. Floyd's triangle是自然数的直角三角形数组 It is defined by filling the rows of the triangle with consecutive numbers, starting with a 1 in the top left corner. 它是通过用连续的数字填充三角…

前言 Hangfire是一个开源且商业免费使用的工具函数库.可以让你非常容易地在ASP.NET应用(也可以不在ASP.NET应用)中执行多种类型的后台任务,而无需自行定制开发和管理基于Windows Service后台任务执行器.且任务信息可以被持久保存.内置提供集成化的控制台. 原文 Hangfire目前资料不多,官方文档提供两个教程 Sending Mail in Background with ASP.NET MVC 和 Highlighter Tutorial,根据第二个教程山寨了一把,…

2588: Spoj 10628. Count on a tree Time Limit: 12 Sec  Memory Limit: 128 MBSubmit: 5217  Solved: 1233[Submit][Status][Discuss] Description 给定一棵N个节点的树,每个点有一个权值,对于M个询问(u,v,k),你需要回答u xor lastans和v这两个节点间第K小的点权.其中lastans是上一个询问的答案,初始为0,即第一个询问的u是明文. Input 第一…

You are given two integer arrays nums1 and nums2 sorted in ascending order and an integer k. Define a pair (u,v) which consists of one element from the first array and one element from the second array. Find the k pairs (u1,v1),(u2,v2) ...(uk,vk) wit…

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below. For example, given the following triangle [ [], [,4], [6,,7], [4,,8,3] ] The minimum path sum from top to bottom is 11 (i.e.,…