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CodeForces Good Bye 2020 A-D

A. Bovine Dilemma 题目分析题意：给你一组数,看他们两两组合的差有多少种情况。那么直接求出两两组合的差，然后去重即可得出答案。去重既可以选择开一个数组标记，也可以选择用STL的unique实现 AC代码 #include<iostream> #include<...

做题记录贪心水题数学思维暴力贡献

2020-12-31

0 676

Educational Codeforces Round 101 B. Red and Blue

题目传送门懒得搬题面了qwq 题目解析 // 解析我也摸鱼了 AC 代码 #include<iostream> #include<algorithm> using namespace std; const int N = 111; int T; bool com(int ...

做题记录数学思维

2020-12-29

0 734

Codeforces Round #690 (Div. 3) C. Unique Number

Codeforces Round #690 (Div. 3) 原题传送门题目描述 You are given a positive number xx. Find the smallest positive integer number that has the sum of digits eq...

做题记录水题思维暴力

2020-12-16

0 813

Codeforces Round #445 B. Vlad and Cafes

Codeforces Round #445 B. Vlad and Cafes [原题传送门](Problem - 890B - Codeforces) 题目描述 Vlad likes to eat in cafes very much. During his life, he has visit...

做题记录水题思维

2020-12-09

0 1144

The 2018 ICPC Asia Nanjing Regional J - Prime Game

The 2018 ACM-ICPC Asia Nanjing Regional Programming Contest J - Prime Game [题目传送门](Attachments - 2018-2019 ACM-ICPC, Asia Nanjing Regional Contest - C...

做题记录思维贡献

2020-12-09

0 963

CodeForeces 1433A

CodeForeces 1433A [题目传送门](Problem - 1433A - Codeforces) 题目描述 There is a building consisting of 10 00010 000 apartments numbered from 11 to 10 00010 0...

做题记录水题

2020-12-08

0 702

2020年广东工业大学第十届文远知行杯新生程序设计竞赛（同步赛）K-很基础的模拟题

K-很基础的模拟题原题传送门题目描述给你一个长度为n的数组，现在定义以下三种操作。 "1 x"表示删除数组的第x(1<=x<=m)个元素，第x+1到第m个元素全部往前移一位。m为操作前数组的元素个数。 "2 x y"表示在数组的第x(1<=x<=m)位插入元素y，原数组...

模拟做题记录水题

2020-12-08

0 795