记$sg(n)={\rm mex}\{sg(n-x)\mid x\in X\}$，考虑如何求$sg(n)$—— 将$m,a,n$均除以$\gcd(m,a)$（其中$n$向下取整），以下假设$m,a$互素 特判$a=0(m=1)$的情况，此时$sg(n)=n$，以下假设$a\in [1,m)$ 用$(i,j)$表示$im+ja$，则$sg(i,j)={\rm mex}\{sg(k,j-1)\mid k\le i\}$（注意$i,j$均可为负）
一、题目大意 标签: 动态归划 https://leetcode.cn/problems/arithmetic-slices 如果一个数列 至少有三个元素 ，并且任意两个相邻元素之差相同，则称该数列为等差数列。 例如，[1,3,5,7,9]、[7,7,7,7] 和 [3,-1,-5,-9] 都是等差数列。 给你一个整数数组 nums ，返回数组 nums 中所有为等
[Arithmetic：Statistical&Statistical：comprehensive] [Arithmetic.Statistical]一、统计学大纲[一.一]：描述统计[一.二]：假设检验/统计推断[一.三]：抽样方法[一.四]：卡方检验[一.五]：t检验、方差检
[Arithmetic：Statistical&Statistical：多指标统计] [Arithmetic.Statistical]二、多指标统计[二.一]：相关分析[二.二]：回归分析[二.三]：因子分析/主成分分析[二.四]：Logistic回归[二.五]：
[Arithmetic：Statistical&Statistical：APP用户数预测] [Arithmetic.Statistical]三、APP用户数预测===============================END===============================Walter Savage Landor:s
题目 An integer array is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same. For example, [1,3,5,7,9], [7,7,7,7], and [3,-1,-5,-9] are arithmetic sequences. Given an integer a
好久没有单独为一道题出题解，但是这道题得确有意义单出一篇题解来做为笔记记录 E. Arithmetic Operations time limit per test 5 seconds memory limit per test 1024 megabytes input standard input output standard output You are given an array of integers
特判\(X \geq 10^9\)的情况 剩下情况暴力枚举 #include <bits/stdc++.h> #define LL long long using namespace std; LL n; int main() { scanf("%lld", &n); if (n >= 1e10) { LL x = n, len = 0; while (x) {len ++; x /= 10;}
In colloquial language, an average is a single number taken as representative of a list of numbers. Different concepts of average are used in different contexts. Often "average" refers to the arithmetic mean, the sum of the numbers divided by ho
词源: count, number, art, craft Arithmetic is a branch of mathematics that consists of the study of numbers, especially the properties of the traditional operations on them - addition, subtraction, multiplication, division, exponentiation and extraction of
HDLBits——Multiplexers Problem 60 : 2-to-1 multiplexer (Mux2to1) Requirement： multiplexer：多路选择器。 本题中需要实现一个 2 选 1 选择器，sel 信号作为选择信号，当 sel = 1 时选择 b，反之选择 a。 Solution： module top_module( input a, b, sel, output out );
Build a 64-bit arithmetic shift register, with synchronous load. The shifter can shift both left and right, and by 1 or 8 bit positions, selected by amount. An arithmetic right shift shifts in the sign bit of the number in the shift register (q[63] in th
原题链接 Problem Description： For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, difference, product and quotient. Input Specification: Each input file contains one test case, which gives in one lin
Overview This course is really hard-core and difficult. We mainly learned about: von Neumann Computer Architecture Arithmetic Unit: Single-Bus, Dual-Bus, Triple-Bus Memory Hierarchy CPU 一、von Neumann Computer Architecture Modern computers are d
// Build a 64-bit arithmetic shift register, // with synchronous load. The shifter can shift both left and right, // and by 1 or 8 bit positions, selected by amount. // An arithmetic right shift shifts in the sign bit of the number in the shift registe
Segment 4：Introduction Number Theory——Arithmetic algorithms【算术算法】：链接 这是整个数论简介的内容，下面是这个的主目录如下，其链接为：https://blog.csdn.net/qq_43479839/article/details/119079620 Segment 4目录@xyi Segment 4：Introduction Number Theory——Arith
原题链接 考察:模拟(?) 思路: 分类讨论就完事. (1) \(n=1\) (2) \(n=2\),这里一定要分\(d = 0\)的情况. (3) \(n>2\),这里一定要注意\(cnt = 0\)的情况,此时合法的情况是只有两种公差,且大公差一定是小公差的\(2\)倍. Code #include <iostream> #include <cstring> #include
为了方便测试，排序方法必须实现该接口。 public interface SortMethod { int[] sortAlgorithm(int[] data); default void swap(int[] data, int i, int j) { int temp = data[i]; data[i] = data[j]; data[j] = temp; } } 排序方法：
数据结构(3) acwing 目录数据结构(3) acwing1.hash表存储结构字符串哈希3.STL1.vector2.string3.queue4.priority_queue(就是堆)5.stack6.deque（双端队列）--basically not use7.set、multiset、map、multimap 1.hash表 哈希函数概念： （1）x mod 10 ^5 即缩小了值域 {模的数最好取
快速链接: . ???????????? 个人博客笔记导读目录(全部) ???????????? 文章目录 Asymmetric Authenticated Encryption Basic Arithmetic Cancellation Converter Data Stream Access Deprecated Events Fast Modular Multiplication Generic Object Generic Operation Initi
牛客网——Rational Arithmetic (20) For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, difference, product and quotient. 输入描述 每个输入文件包含一个测试用例，它在一行中给出了格式为“a1/b1 a2/b2”
PAT 甲级 1088 Rational Arithmetic 大家注意一下Inf不要写成INF不然就会像个憨批一样扒数据扒半个小时。 #include <bits/stdc++.h> using namespace std; struct Fraction { long long up, down; }; long long gcd(long long a, long long b) { if (b == 0) return
https://codeforces.com/problemset/problem/382/C 题意:给定一个序列，问是否可以通过只插入一个数来使得整个序列成为等差数列，求出总共有多少可能的情况，并输出这些数。 思路: 模拟一下发现只有几种情况..然后要考虑各种细节。 其中一个是两个不同的差出现次数相等的时候要用大
[CF1114E] Arithmetic Progression Description 有一个长度为 n 的序列 a，从小到大排序后是一个等差数列。? i 询问 ai 的值，> x 询问序列中是否存在严格大于 x 的数，要求出首项和公差。询问个数不超过 60。 Solution 用第二种操作二分求出序列最大值，再用第一种操作随机询问若干个数，
