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  • Codeforces Round #624 (Div. 3) (A~D,CD Good)2021-08-26 17:05:31

    比赛链接:Here 1311A. Add Odd or Subtract Even 签到题, \(a > b\) 时必须做做减法,如果差值为偶数的话只需要 \(1\) 次不然做一次减法后再做一次 \(+1\) 即可 \(a < b\) 同理了 \(a = b\) 0次 int main() { cin.tie(nullptr)->sync_with_stdio(false); int _; for (cin >

  • 【疾病分类】基于matlab农作物叶子虫害识别与分类【含Matlab源码 624期】2021-07-05 11:06:19

    一、简介 基于matlab农作物叶子虫害程序识别 二、源代码 clear all clc disp('正在训练农作物叶子图像模板,请稍后...'); disp(' '); %color_Ip = xunlian(); pause(2); load C:\Users\lenovo\Desktop\图像检索\color_Ip.mat; disp('图像训练完成,正在进行图像识别,请稍后...'); d

  • 【20210626GKCTF】不想考试又是混混子的一天2021-07-04 14:29:31

    GKCTF 26号,那天考操作系统,没有参加;27下午考完计网,想要复现一下 文章目录 GKCTFCryptorandom题目描述解题思路脚本搜集失败的思路 ReverseQQQQT 其他战队的WP Crypto BUU已经上题 这次的应急杯Crypto有三道,Reverse有五道。根据解出的人数,今日任务就复现下两道解出人数

  • 【DB笔试面试624】在Oracle中,Hash Join是不是有排序?Hash Join会在什么时候慢?2021-04-15 07:06:27

    ♣          题目         部分 在Oracle中,Hash Join是不是有排序?Hash Join会在什么时候慢?     ♣          答案部分          哈希连接(Hash Join,HJ)自身不需要排序,这是区别排序合并连接(Sort Merge Join,SMJ)的特点之一。Hash Join原理比较复杂,但是如

  • Codeforces Round #624 (Div. 3) problem C2020-02-28 19:44:20

    C. Perform the Combo time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output You want to perform the combo on your opponent in one popular fighting game. The combo is the string s consisting of n lowerca

  • Codeforces #624 div3 C2020-02-28 19:01:07

    You want to perform the combo on your opponent in one popular fighting game. The combo is the string ss consisting of nn lowercase Latin letters. To perform the combo, you have to press all buttons in the order they appear in ss. I.e. if s=s="abca&qu

  • Codeforces Round #624 (Div. 3) A. Add Odd or Subtract Even(水题)2020-02-27 15:03:20

    You are given two positive integers aa and bb . In one move, you can change aa in the following way: Choose any positive odd integer xx (x>0x>0 ) and replace aa with a+xa+x ; choose any positive even integer yy (y>0y>0 ) and replace aa with

  • Codeforces Round #624 (Div. 3)简要题解2020-02-27 13:09:53

    A. Add Odd or Subtract Even 题意:给两个数xy,每次对x加上一个奇数或者减去一个偶数,问从x变到y的最小操作次数。 分析:判断一下x和y的奇偶性讨论一下就可以了。 #include <bits/stdc++.h> using namespace std; int main(){ int t; cin>>t; while(t--){ int

  • Codeforces Round #624 (Div. 3)2020-02-25 19:01:10

    挺简单的 题目链接:https://codeforces.com/contest/1311 A: 白给 1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define pb emplace_back 6 #define mp make_pair 7 #define eps 1e-8 8 #define lson (curpos<<1)

  • Codeforces Round #624 (Div. 3)C. Perform the Combo2020-02-25 18:41:59

    You want to perform the combo on your opponent in one popular fighting game. The combo is the string s consisting of n lowercase Latin letters. To perform the combo, you have to press all buttons in the order they appear in s. I.e. if s=“abca” then yo

  • Codeforces Round #624 (Div. 3)2020-02-25 18:37:57

    A 选正奇数 增加,正奇数减少,首先判断两个数的大小关系。看是该增加还是减少,然后答案显然只有0,1,2。 int main(){ int t = read(); while(t--){ ll a,b; a = read();b = read(); if(a == b) puts("0"); else if(a < b){

  • Codeforces Round #624 (Div. 3) D 题2020-02-25 17:03:54

    题目链接:Three Integers 简要题意: 给出 \(a, b, c\) 满足 \(a \leq b \leq c\)。 你可以随便选择其中一个数,使这个数 \(+1\) 或 \(-1\)。 假设最后操作完 \(a\) 为 \(A\),\(b\) 为 \(B\),\(c\) 为 \(C\)。 你需要使 \(A | B\) 且 \(B | C\),并最小化操作次数。 输出操作次数和 \(A,

  • UVA-6242020-01-14 20:08:13

    题目 You have a long drive by car ahead. You have a tape recorder, but unfortunately your best music is on CDs. You need to have it on tapes so the problem to solve is: you have a tape N minutes long. How to choose tracks from CD to get most out of tape

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