Codeforces Round #542(Div. 2) D1.Toy Train】的更多相关文章

链接:https://codeforces.com/contest/1130/problem/D1 题意: 给n个车站练成圈,给m个糖果,在车站上,要被运往某个位置,每到一个车站只能装一个糖果. 求从每个位置开车的最小的时间. 思路: vector记录每个位置运送完拥有糖果的时间消耗,为糖果数-1 * n 加上消耗最少时间的糖果. 对每个起点进行运算,取所有点中的最大值. 代码: #include <bits/stdc++.h> using namespace std; typedef lon…
layout: post title: Codeforces Round 542 (Div. 2) author: "luowentaoaa" catalog: true tags: mathjax: true - codeforces - 并查集 --- 传送门 前三题太简单不写 D.Toy Train (贪心) 题意 有n个车站,按照环前进,有m条要求从x送到y,每次从x最多能拿一个糖,输出在第i个车站出发最少需要多少时间完成所有要求 (注意车的容量无穷) 思路 所以我们直接枚举每…
题目传送门 /* 题意:5种情况对应对应第i或j辆车翻了没 水题:其实就看对角线的上半边就可以了,vis判断,可惜WA了一次 3: if both cars turned over during the collision. 是指i,j两辆车,而不是全部 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <iostream>…
D1. Magic Powder - 1 题目连接: http://www.codeforces.com/contest/670/problem/D1 Description This problem is given in two versions that differ only by constraints. If you can solve this problem in large constraints, then you can just write a single soluti…
任意门:http://codeforces.com/contest/1118/problem/D1 D1. Coffee and Coursework (Easy version) time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The only difference between easy and hard versions…
传送门:http://codeforces.com/contest/1092/problem/D1 D1. Great Vova Wall (Version 1) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vova's family is building the Great Vova Wall (named by Vo…
 A. Toy Cars Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/545/problem/A Description Little Susie, thanks to her older brother, likes to play with cars. Today she decided to set up a tournament between them. The process…
 传送门 Description Little Chris is very keen on his toy blocks. His teacher, however, wants Chris to solve more problems, so he decided to play a trick on Chris. There are exactly s blocks in Chris's set, each block has a unique number from 1 to s. Chr…
题目链接: 题目 D. Toy Sum time limit per test:1 second memory limit per test:256 megabytes 问题描述 Little Chris is very keen on his toy blocks. His teacher, however, wants Chris to solve more problems, so he decided to play a trick on Chris. There are exactly…
C https://codeforces.com/contest/1130/problem/C 题意 给你一个\(n*m\)(n,m<=50)的矩阵,每个格子代表海或者陆地,给出在陆地上的起点终点,只允许挖一条穿越海的隧道,假设隧道连接的两个陆地分别为(x1,y1),(x2,y2),则挖隧道的花费为\((y1-x1)*(y1-x1)+(y2-x2)*(y2-x2)\) 题解 分别找出和起点,和终点连接的陆地(找联通块),枚举两个点维护最小值 代码 #include<bits/stdc++.h&…
D. Toy Sum   time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Little Chris is very keen on his toy blocks. His teacher, however, wants Chris to solve more problems, so he decided to play a t…
D1. Magic Powder - 1 time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output This problem is given in two versions that differ only by constraints. If you can solve this problem in large constraint…
链接:https://codeforces.com/contest/1130/problem/B 题意: 给定n和 2 * n个数,表示i位置卖ai层蛋糕, 有两个人在1号,必须严格按照1-n的顺序买蛋糕,同时每个店只买一个蛋糕 . 求所需的最短时间. 思路: 将每种蛋糕对应位置记录在二维数组. 从1-n挨个买,根据上一次的位置算出消耗. 代码: #include <bits/stdc++.h> using namespace std; typedef long long LL; const…
链接:https://codeforces.com/contest/1130/problem/A 题意: 给n个数,找出一个非0整数d,使所有n个数除以整数d后,数组中正数的数量>= n/2. 如果不存在d,输出0. 思路: 记录正数和负数的数量,只有有一个满足就输出1或-1,否则为0. 代码: #include <bits/stdc++.h> using namespace std; typedef long long LL; int main() { int n, a; cin &g…
链接:https://codeforces.com/contest/1130/problem/C 题意: 给一个n*n的图,0表示地面,1表示水,给出起点和终点, 现要从起点到达终点,有一次在两个坐标间创立隧道的机会,消耗为(x1 - x2)^2 + (y1 - y1)^2. 求出最小的消耗,如果直接能走到,则消耗为0. 思路: bfs求出起点和终点分别能到达的点,枚举每一种情况,取最小值即可. 代码: #include <bits/stdc++.h> using namespace std;…
D1. RGB Substring (easy version) time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output The only difference between easy and hard versions is the size of the input. You are given a string s consistin…
一.题目 D1. Submarine in the Rybinsk Sea (easy edition) 二.分析 简单版本的话,因为给定的a的长度都是定的,那么我们就无需去考虑其他的,只用计算ai的值在每个位置的贡献即可. 因为长度是定的,如果ai在前,那么对所有的a的贡献就是在偶数位的贡献值然后乘以n即可. 如果ai在后,那么对所有ai的贡献就是在奇数位的贡献值然后乘以n. 将两种情况合并,其实就是求ai在每个位置下的贡献,然后乘以n. 时间复杂度是$O(n)$ 三.AC代码 1 #incl…
开学了住校了打不了深夜场 好难受啊QwQ A 显然对于每个起点,我们只需要贪心记录这个起点出发出去的糖果数量以及离自己最近的糖果 因为这个起点最后一次装载糖果一定是装载终点离自己最近的那个糖果 $ O(n^2)$暴力贪心即可 有线性做法懒得写了 #include<ctime> #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algo…
A:显然对于起点相同的糖果,应该按终点距离从大到小运.排个序对每个起点取max即可.读题花了一年还wa一发,自闭了. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N…
思路:枚举换的位置i,j 然后我们要先判断改序列能否完全匹配 如果可以 那我们就需要把差值最大的位置换过来 然后直接判断就行…
思路:枚举换的位置i,j 然后我们要先判断改序列能否完全匹配 如果可以 那我们就需要把差值最大的位置换过来 然后直接判断就行…
Codeforces Round #542 [Alex Lopashev Thanks-Round] (Div. 2) 题目链接:https://codeforces.com/contest/1130 A. Be Positive 题意: 给出n个数,看是否正数的个数过半或者负数的个数过半. 题解:水题,代码如下: #include <bits/stdc++.h> using namespace std; ; int n; double a[N]; int main(){ ios::sync_…
Codeforces Round #258 (Div. 2)[ABCD] ACM 题目地址:Codeforces Round #258 (Div. 2) A - Game With Sticks 题意:  Akshat and Malvika两人玩一个游戏,横竖n,m根木棒排成#型,每次取走一个交点,交点相关的横竖两条木棒要去掉,Akshat先手,给出n,m问谁赢. 分析:  水题,非常明显无论拿掉哪个点剩下的都是(n-1,m-1),最后状态是(0,x)或(x,0),也就是拿了min(n,m)-…
今天老师(orz sansirowaltz)让我们做了很久之前的一场Codeforces Round #257 (Div. 1),这里给出A~C的题解,对应DIV2的C~E. A.Jzzhu and Chocolate time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has a big rectangular cho…
Codeforces Round #551 (Div. 2) 算是放弃颓废决定好好打比赛好好刷题的开始吧 A. Serval and Bus 处理每个巴士最早到站且大于t的时间 #include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ')…
Codeforces Round #515 (Div. 3) #include<bits/stdc++.h> #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> #include<queue> #include<vector> #incl…
Codeforces Round #552 (Div. 3) 题目链接 A. Restoring Three Numbers 给出 \(a+b\),\(b+c\),\(a+c\) 以及 \(a+b+c\) 这四个数,输出一种合法的 \(a,b,c\).   可以发现,前面的两个数加起来减去最后的 \(a+b+c\),答案就出来一个.最后这样求出\(a,b,c\)即可. 代码如下: Code #include <bits/stdc++.h> using namespace std; typede…
//在我对着D题发呆的时候,柴神秒掉了D题并说:这个D感觉比C题简单呀!,,我:[哭.jpg](逃 Codeforces Round #435 (Div. 2) codeforces 862 A. Mahmoud and Ehab and the MEX[水] 题意:定义一个集合的MEX为其中的最小的不在集合里的非负整数,现在给一个大小为N的集合和X,每次操作可以向其中加入一个非负整数或删除一个数,问最小的操作次数,使得这个集合的MEX为X. #include<cstdio> #include…
Codeforces Round #527 (Div. 3) 题解 题目总链接:https://codeforces.com/contest/1092 A. Uniform String 题意: 输入n,k,n表示字符串的长度,k表示从1-k的小写字符(1即是a),现在要求最大化最少字符的数量. 题解: 贪心搞一搞就行了. 代码如下: #include <bits/stdc++.h> using namespace std; int T; int n,k; int main(){ cin>…
题目链接:Codeforces Round #298 (Div. 2) A. Exam An exam for n students will take place in a long and narrow room, so the students will sit in a line in some order. The teacher suspects that students with adjacent numbers (i and i + 1) always studied side…