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题解 CF1575D【Divisible by Twenty-Five】
值域非常小，其中只有 \(4\times 10^6\) 个数是 \(25\) 的倍数，因此可以暴力枚举所有位数正确的 \(25\) 的倍数，然后检查是否合法。 检查方法就是枚举每一位，如果是数字就必须一模一样，否则记录字符 X 表示的数是多少，看看是否自始至终都是一样的。 需要特判 \(|s|=1\) 的情况。 // ProblARC141D Non-divisible Set
题面传送门 神仙题，被踩爆了/kk 首先证明一个结论：\(\{1,2,\dots 2m\}\)中，按照这样的最大独立集，答案为\(m\) 首先证明下界，显然是\(\{m+1,m+2,\dots 2m\}\)这样的一组合法解。 然后证明上界，我们将每个数的\(2\)都除掉，就只剩下\(m\)个数，这\(m\)个数中显然不能有两个同组的，所以答案上限&& || ^
import java.util.*; import javax.swing.*; class Main { public static void main(String[] args){ Scanner input = new Scanner(System.in); System.out.print("Enter an integer: "); int number = input.nextInt();[CodeForces] The Meaningless Game
Problem S: the set of numbers said by player 1; P: the set of numbers said by player 2; S^2 * P = a; S * P^2 = b; a * b = (S * P)^3, so a * b must be a perfect cubic number. Binary search to check this. Let X be the cubic root of a * b, X = S * P. In oB. Make it Divisible by 25(900)
#include<bits/stdc++.h> using namespace std; int main(){ int n; string a; cin>>n; for(int i=1;i<=n;i++){ cin>>a; int minn=100; int len=a.size(); for(int j=0;j<len;j++){ fo[LeetCode]1262. Greatest Sum Divisible by Three (JS) 动态规划&小学数学解法
题目描述 LeetCode原题链接：1262. Greatest Sum Divisible by Three Given an array nums of integers, we need to find the maximum possible sum of elements of the array such that it is divisible by three. Example 1: Input: nums = [3,6,5,1,8] Output: 18 Explanation【leetcode】1010. Pairs of Songs With Total Durations Divisible by 60
You are given a list of songs where the ith song has a duration of time[i] seconds. Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i, j such that i < j with【leetcode】1015. Smallest Integer Divisible by K
Given a positive integer k, you need to find the length of the smallest positive integer n such that n is divisible by k, and n only contains the digit 1.Return the length of n. If there is no such n, return -1. Note: n may not fit inLeetCode --- 1018. Binary Prefix Divisible By 5 解题报告
You are given a binary array nums (0-indexed). We define xi as the number whose binary representation is the subarray nums[0..i] (from most-significant-bit to least-significant-bit). For example, if nums = [1,0,1], then x0 = 1, x1 = 2, and x2【Rust】函数
环境 Rust 1.56.1 VSCode 1.61.2 概念 参考：https://doc.rust-lang.org/stable/rust-by-example/fn.html 示例 定义函数 使用关键字 fn 定义函数，函数定义和顺序无关。 fn main() { say() } // main 函数虽然在 say 的前面定义，但还是可以使用。 fn say() { println!("helSolution -「营业」「ABC 170」Not Divisible
Description Link. 给出一个长度为 \(n\) 的序列 \(a\)，求 \(\sum_{i=1}^{n}[\forall j\in[1,i)\cup(i,n],a_{j}\nmid a_{i}]\)。 Solution 首先特判序列中有 \(1\) 的情况。 然后调和级数把每个数的倍数开桶记录。 最后扫一遍序列，看该元素在桶里面出现的次数不超过一就有贡献。 #iMatlab中continue语句
目录 一.语法 二.说明 三.示例 1.有选择地显示循环中的值 2.跳至下一循环迭代 将控制权传递给 for 或 while 循环的下一迭代。 一.语法 continue 提示 continue 语句跳过 for 或 while 循环中剩余的说明，并开始下一迭代。要完全退出循环，请使用 break 语句。 conti【leetcode】368. 最大整除子集（largest-divisible-subset）（模拟）[DP]
链接 https://leetcode-cn.com/problems/largest-divisible-subset/ 耗时 解题：46 min 题解：19 min 题意 给你一个由 无重复 正整数组成的集合 nums ，请你找出并返回其中最大的整除子集 answer ，子集中每一元素对 (answer[i], answer[j]) 都应当满足： answer[i] % answer[j] == 0Codeforces Round #708 (Div. 2) B. M-arrays 思维+哈希表
原题链接 测试样例 input 4 6 4 2 2 8 6 9 4 10 8 1 1 1 5 2 4 4 8 6 7 1 1 666 2 2 2 4 output 3 6 1 1 Note In the first test case we can divide the elements as follows: [ 40526. Beautiful Arrangement (M)
Beautiful Arrangement (M) 题目 Suppose you have n integers from 1 to n. We define a beautiful arrangement as an array that is constructed by these n numbers successfully if one of the following is true for the ith position (1 <= i <= n) in this array:[LeetCode] 974. Subarray Sums Divisible by K 子数组数字之和可被K整除
Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K. Example 1: Input: A = [4,5,0,-2,-3,1], K = 5 Output: 7 Explanation: There are 7 subarrays with a sum divisible by K = 5: [4, 5, 0, -2, -3,leetcode学习笔记1010 Pairs of Songs With Total Durations Divisible by 60
leetcode学习笔记1010 开始之前问题方法1 开始之前 问题 Pairs of Songs With Total Durations Divisible by 60 You are given a list of songs where the ith song has a duration of time[i] seconds. Return the number of pairs of songs for which their total du[LeetCode] 1010. Pairs of Songs With Total Durations Divisible by 60
In a list of songs, the i-th song has a duration of time[i] seconds. Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i, j such that i < j with (time[i]【leetcode_easy_array】1010. Pairs of Songs With Total Durations Divisible by 60
problem 1010. Pairs of Songs With Total Durations Divisible by 60 参考 1. leetcode_1010. Pairs of Songs With Total Durations Divisible by 60； 完Python_有返回值的函数_布尔函数
函数也可以返回布尔值(True或False)，这种情况便于隐藏函数内部的复杂测试。例如: #!/bin/python def is_divisible(x, y): if x % y == 0: return True else: return False print is_divisible(6, 3) $ python a.py True 一般情况下都给这[LeetCode] 368. Largest Divisible Subset
Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj) of elements in this subset satisfies: Si % Sj = 0 or Sj % Si = 0. If there are multiple solutions, return any subset is fine. Example 1: Input: [1,2,3] Ou1010. Pairs of Songs With Total Durations Divisible by 60
问题： 给出一列歌曲所花时间的数组。 求任意两首歌合起来时间是60分钟的倍数的组队pair数。 Example 1: Input: [30,20,150,100,40] Output: 3 Explanation: Three pairs have a total duration divisible by 60: (time[0] = 30, time[2] = 150): total duration 180 (time[1] = 2Leap year
Leap year is defined as a year that divisible by 400 or not divisible by 100 but divisible by 4. #include <iostream>using namespace std; int main(){ int y; cin>>y; if((y%4==0&&y%100!=0)||y%400==0) cout<<"yes&quo埃氏筛法求素数
计算素数的一个方法是埃氏筛法，它的算法理解起来非常简单： 首先，列出从2开始的所有自然数，构造一个序列： 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, ... 取序列的第一个数2，它一定是素数，然后用2把序列的2的倍数筛掉： 3, 4, 5, 6, 7, 8, 9, 10, 11, 1LeetCode:1010. Pairs of Songs With Total Durations Divisible by 60 总持续时间可被 60 整除的歌曲(C语言)
题目描述： 在歌曲列表中，第 i 首歌曲的持续时间为 time[i] 秒。 返回其总持续时间（以秒为单位）可被 60 整除的歌曲对的数量。形式上，我们希望索引的数字 i < j 且有 (time[i] + time[j]) % 60 == 0。 示例 1： 输入：[30,20,150,100,40] 输出：3 解释：这三对的总持续时间可被 60 整数： (