题目链接:Codeforces Round #367 (Div. 2) D. Vasiliy's Multiset 题意: 给你一些操作,往一个集合插入和删除一些数,然后?x让你找出与x异或后的最大值 题解: trie树xjb搞就行,每次要贪心,尽量满足高位为1. #include<bits/stdc++.h> #define F(i,a,b) for(int i=a;i<=b;i++) using namespace std; namespace trie { )*; ],ed=-,c…

Vasiliy's Multiset 题目链接: http://codeforces.com/contest/706/problem/D Description Author has gone out of the stories about Vasiliy, so here is just a formal task description. You are given q queries and a multiset A, initially containing only integer…

题目链接: http://codeforces.com/contest/706/problem/D D. Vasiliy's Multiset time limit per test:4 secondsmemory limit per test:256 megabytes 问题描述 Author has gone out of the stories about Vasiliy, so here is just a formal task description. You are given q…

D. Vasiliy's Multiset time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Author has gone out of the stories about Vasiliy, so here is just a formal task description. You are given q queries a…

D. Vasiliy's Multiset time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Author has gone out of the stories about Vasiliy, so here is just a formal task description. You are given q queries a…

http://codeforces.com/contest/706/problem/D 题意:有多种操作,操作1为在字典中加入x这个数,操作2为从字典中删除x这个数,操作3为从字典中找出一个数使得与给定的数的异或值最大. 思路: 因为这道题目涉及到删除操作,所以用一个变量cnt来记录前缀的数量,加入时就+1,删除时就减1.查询时前缀数量>0时就说明是存在的. #include<iostream> #include<cstdio> #include<cstring>…

D. Vasiliy's Multiset time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Author has gone out of the stories about Vasiliy, so here is just a formal task description. You are given q queries a…

http://codeforces.com/group/1EzrFFyOc0/contest/706/problem/D 题目:就是有3种操作 + x向集合里添加 x - x 删除x元素,(保证存在 ? x 查询 x |  集合中元素的最大值 思路:就是利用字典树,从高位到低位进行贪心. 比如说给一个数 x=3  , 对x 各位取反(二进制)(x=~x ), 于是就是 1-0-0-1: 拿 1-0-0-1,从左到右(从高位到地位)顺序,来在字典树中寻找.如果能找到(if ),就接着找下去: 如果…

E. Ann and Half-Palindrome time limit per test 1.5 seconds memory limit per test 512 megabytes input standard input output standard output Tomorrow Ann takes the hardest exam of programming where she should get an excellent mark. On the last theoreti…

A.Beru-taxi 水题:有一个人站在(sx,sy)的位置,有n辆出租车,正向这个人匀速赶来,每个出租车的位置是(xi, yi) 速度是 Vi;求人最少需要等的时间: 单间循环即可: #include<iostream> #include<algorithm> #include<string.h> #include<stdio.h> #include<math.h> #include<vector> using namespace…

Hard problem 题目链接: http://codeforces.com/contest/706/problem/C Description Vasiliy is fond of solving different tasks. Today he found one he wasn't able to solve himself, so he asks you to help. Vasiliy is given n strings consisting of lowercase Engl…

Interesting drink 题目链接: http://codeforces.com/contest/706/problem/B Description Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bough…

Beru-taxi 题目链接: http://codeforces.com/contest/706/problem/A Description Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi…

Codeforces Round 367 Div. 2 点击打开链接 A. Beru-taxi (1s, 256MB) 题目大意:在平面上 \(n\) 个点 \((x_i,y_i)\) 上有出租车,每辆出租车的行驶速度为 \(v_i\),求所有出租车到点 \((a,b)\) 的时间中的最短时间. 数据范围:\(-100 \leq a,b,x_i,y_i \leq 100\),\(v_i \leq 100\),\(n \leq 1000\) 简要题解:依次求出所有的时间,取最小值即可. 时空复杂度…

题目链接:Codeforces Round #367 (Div. 2) C. Hard problem 题意: 给你一些字符串,字符串可以倒置,如果要倒置,就会消耗vi的能量,问你花最少的能量将这些字符串排成字典序 题解: 当时1点过头太晕了,看错题了,然后感觉全世界都会,就我不会,- -!结果就是一个简单的DP, 设dp[i][0]表示第i个字符串不反转的情况,dp[i][1]表示第i个字符串反转的情况 状态转移方程看代码 #include<bits/stdc++.h> #define F(…

A题 Beru-taxi 随便搞搞.. #include <cstdio> #include <cmath> using namespace std; int a,b,n; struct _ { int x,y,v; }p[]; double dis2(_ A) { return (a - A.x) * (a - A.x) + (b - A.y) * (b - A.y); } int main() { scanf("%d%d%d", &a, &b…

吐槽:只能说是上分好场,可惜没打,唉 A:Beru-taxi (水题,取最小值) #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string>…

https://codeforces.com/contest/1111/problem/D 多重排列 + 反向01背包 题意: 给你一个字符串(n<=1e5,n为偶数),有q个询问,每次询问两个位置x和y,问将和x,y相同的所有字符移到前半段或者后半段,并且剩下的所有字符都要在同一半段的方案数,字符是大写小写字母 题解: 首先不考虑x和y位置,假设前半段的排列数为\((n/2)!/(cnt[x_1]!*...*cnt[x_n]!)\),后半段的排列为\((n/2)!/(cnt[y_1]!*...…

A. Beru-taxi time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as…

A. Beru-taxi time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as…

Hard problem 题意: 有n个字符串,对第i个字符串进行反转操作代价为ci. 要使n个字符串按照字典序从小到大排列,最小的代价是多少. 题解: 反转就是reverse操作,比如说45873反转之后只能是37845,不能是别的,当时就这没有理解好,所以没继续去想,其实可以假设这样,之后来一发的,如果当时这样不就对了嘛. 一行只有反转或不反转,要求最小代价,有道01背包的意思,所以就要dp试试,dp[i]代表第i的串的最小代价,但是怎么继续推呢?,没有办法,就再多开一维,另一维只要代表2个…

题意与分析 题意:给出\(n\)个字符串,可以反转任意串,反转每个串都有其对应的花费\(c_i\).经过操作后是否能满足字符串\(\forall i \in [1,n] \text{且} i \in R_+, str[i]\ge str[i-1]\),若能输出最小花费,否则输出-1. 分析:经过各种字符串dp血虐,应该会有个直觉:\(dp[i]\)表示前\(i\)个串的最小花费.但是好像不太够:没有保存反转.因此,在dp中,如果状态不够,那就加维度保存状态.这里就是:我们定义\(dp[i][0]…

题目链接:http://codeforces.com/contest/703/problem/D 思路:看了神犇的代码写的... 偶数个相同的数异或结果为0,所以区间ans[l , r]=区间[l , r]每个数相异或^区间[l , r]出现过的数相异或.如数组1,2,1,3,3,2,3,则ans[1 , 7]=(1^2^1^3^3^2^3)^(1^2^3) 前半部分可以处理出前缀异或,后半部分先对询问按r进行排序,处理过程中相同的数只保留最后一个. #include<bits/stdc++.h…

题目链接:http://codeforces.com/contest/703/problem/D 给你n个数,m次查询,每次查询问你l到r之间出现偶数次的数字xor和是多少. 我们可以先预处理前缀和Xor[i],表示1~i的xor和.因为num^num=0,所以Xor[r] ^ Xor[l - 1]求的是l~r之间出现奇数次的数字xor和. 那怎么求偶数次的呢,那我们可以先求l到r之间不重复出现数字的xor(比如1 1 2 求的是1 ^ 2),然后再xor以上求出的Xor[r] ^ Xor[l…

题目链接:http://codeforces.com/contest/474/problem/F 题意简而言之就是问你区间l到r之间有多少个数能整除区间内除了这个数的其他的数,然后区间长度减去数的个数就是答案. 要是符合条件的话,那这个数的大小一定是等于gcd(a[l]...a[r]). 我们求区间gcd的话,既可以利用线段树性质区间递归下去然后返回求解,但是每次查询是log的,所以还可以用RMQ,查询就变成O(1)了. 然后求解区间内有多少个数的大小等于gcd的话,也是利用线段树的性质,区间递…

题目链接:http://codeforces.com/problemset/problem/438/D 给你n个数,m个操作,1操作是查询l到r之间的和,2操作是将l到r之间大于等于x的数xor于x,3操作是将下标为k的数变为x. 注意成段更新的时候,遇到一个区间的最大值还小于x的话就停止更新. #include <iostream> #include <cstdio> #include <cstring> using namespace std; typedef __…

C. Day at the Beach Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/599/problem/C Description One day Squidward, Spongebob and Patrick decided to go to the beach. Unfortunately, the weather was bad, so the friends were unab…

D. The Child and Sequence Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/438/problem/D Description At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important…

B. "Or" Game Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/578/problem/B Description You are given n numbers a1, a2, ..., an. You can perform at most k operations. For each operation you can multiply one of the number…

题目地址:http://codeforces.com/contest/474/problem/F 由题意可知,最后能够留下来的一定是区间最小gcd. 那就转化成了该区间内与区间最小gcd数相等的个数.区间最小gcd一定小于等于区间最小值.所以仅仅要先推断最小值是否是最小gcd.若是的话,就求出最小值的个数.然后用r-l+1-个数就可以. 对于以上信息.能够用线段树来维护.分别维护区间gcd,区间最小值以及区间最小值的个数. 代码例如以下: #include <iostream> #includ…