题目 题意:一个人可以在一分钟同时进行m道菜的一个步骤,共有n道菜,每道菜各有xi个步骤,求做完的最短时间. 思路:一道很水的思维题, 根本不需要去 考虑模拟过程 以及先做那道菜(比赛的时候就是这么考虑的). 只是需要判断总数的平均值 和 耗时最大的一道菜 哪个最大.. #include <iostream> #include <cstdio> #include <cstring> #include <cstring> #include <cmath&…

题目链接 2014年浙江省赛C题,当时觉得难,现在想想这题真水.. 找规律: 若   最大的那个步骤数*m-总和>=0,那么答案就是 最大的那个步骤数 . 否则  就要另加上不够的数量,具体看代码吧,嘻嘻. 下面这个是我比赛时写的,紧张时写的有点冗杂,开心的是一次过了,哈哈. 数组dp[i]是装逼的,保存的是前i个所需的最少时间,貌似除了dp[n-1],前面的都是多余的 - - . #include<stdio.h> #include<string.h> int main()…

题目链接 题意 : 这个人需要做n道菜,每道菜Ai步,他可以同时做M道不同的菜的其中一步,问你最少需要多少时间能做完所有的菜. 思路 : 这个题比赛的时候禁锢思路了,根本没想出来,就是当M > N时,需要的时间一定是N道菜里边步骤最多的那道菜的步骤数.如果不是就判断平均需要的时间,如果平均需要的时间不如最多步骤多,那还是步骤数,否则就是平均时间. #include <iostream> #include <stdio.h> #include <string.h>…

这题的意思是给你 n 道菜,第 i 道菜需要 Ai 步才能完成 每次你能对 m 道菜分别完成一步,请问最少需要几次? 这题暴力写肯定是不行的,去年省赛的时候就是没写出来这题,今天再把思路理一理吧. 首先我们需要明确的是 1. n <= m 的时候, 那么答案显而易见,就是 Max (A[i]) 2. n > m ①此时我们不难发现这个现象,如果 sum (A[i]) 能被 m 整除,那么我们可能恰恰只需要 (sum / m) 次即可完成 一个例子: n = 3, m = 2 a[1] = 1,…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3778 题目 某人做菜很厉害,一分钟能同时完成最多m个菜的一道工序,输入菜的个数n和整数m,再输入每个菜的工序数,输出最短时间. 范围: 1 <= N, M <= 40000 1 <= Ai <= 40000 Sample Input 2 3 2 2 2 2 10 6 1 2 3 4 5 6 7 8 9 10 Sample Output 3 10 题解 贪心,…

Talented Chef Time Limit: 2 Seconds Memory Limit: 65536 KB As we all know, Coach Gao is a talented chef, because he is able to cook M dishes in the same time. Tonight he is going to have a hearty dinner with his girlfriend at his home. Of course, Coa…

As we all know, Coach Gao is a talented chef, because he is able to cook M dishes in the same time. Tonight he is going to have a hearty dinner with his girlfriend at his home. Of course, Coach Gao is going to cook all dishes himself, in order to sho…

Description As we all know, Coach Gao is a talented chef, because he is able to cook M dishes in the same time. Tonight he is going to have a hearty dinner with his girlfriend at his home. Of course, Coach Gao is going to cook all dishes himself, in…

As we all know, Coach Gao is a talented chef, because he is able to cook M dishes in the same time. Tonight he is going to have a hearty dinner with his girlfriend at his home. Of course, Coach Gao is going to cook all dishes himself, in order to sho…

题目链接 TAG: 这是我近期做过最棒的一道贪心思维题,不容易想到,想到就出乎意料. 题意:给定两个含有N个正整数的数组a和b,让你输出一个数字k ,要求k不大于n/2+1,并且输出k个整数,范围为1~n的不重复数字, 要求这k个数字为下标的对应a和b中的数的和乘以2的值  分别大于a和b 的数组总和. 思路:首先对a进行降序排序,然后输出最大值的下标,随后进行幅度为2的枚举,对排序后的a2~an进行选择性输出下标,(注意,排序的时候用一个新数组开两个变量,一个index,一个v进行排序,可以用…

https://codeforces.com/contest/1136/problem/D 贪心 + 思维 题意 你面前有一个队列,加上你有n个人(n<=3e5),有m(m<=个交换法则,假如u在v相邻前面,那么u和v可以交换位置,问你是队列最后一个人的时候你最前可以换到前面哪里 题解 因为相邻才能换,所以最后一个换到前面一定是一步一步向前走,所以不存在还要向后走的情况 设最后一个为u,假设前面有一个能和u换位置的集合,那么需要将这些点尽量往后移动去接u 假设前面有一个不能和u换位置的集合S,…

题目传送门 /* 题意:套娃娃,可以套一个单独的娃娃,或者把最后面的娃娃取出,最后使得0-1-2-...-(n-1),问最少要几步 贪心/思维题:娃娃的状态:取出+套上(2),套上(1), 已套上(0),先从1开始找到已经套好的娃娃层数, 其他是2次操作,还要减去k-1个娃娃是只要套上就可以 详细解释:http://blog.csdn.net/firstlucker/article/details/46671251 */ #include <cstdio> #include <algor…

题目传送门 /* 题意:n个头,m个士兵,问能否砍掉n个头 贪心/思维题:两个数组升序排序,用最弱的士兵砍掉当前的头 */ #include <cstdio> #include <cstring> #include <algorithm> using namespace std; ; const int INF = 0x3f3f3f3f; int a[MAXN], b[MAXN]; int main(void) //UVA 11292 The Dragon of Loo…

Description Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter. Their algorithm will be tested on an array of integers, where the i-th integer represents the…

LINK:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3778 题意:有n道菜,每道菜需要\(a_i\)道工序,有m个锅可以同时做不同菜,问最小时间. 思路:水题,但要注意最小需要的时间显然是最大的\(a_i\)值,剩下的就是求个均,求个余的事了. /** @Date : 2017-03-24-14.35 * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : http…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3829 现场做这道题的时候,感觉是思维题.自己智商不够.不敢搞,想着队友智商好,他们搞吧.可是没出来这题...... 以后不论什么时候,都自信点....该想的还是好好自己想,这类题感觉就是先去找性质,然后一点点找规律,假设必要的话.自己提出一点猜想.然后假设自己举不出来反例,就临时觉得是正确的 下午搞了一下午.发现还是悲剧,晚上參考了两个题解 http://blog.csd…

C. Coffee Break 这个贪心之前好像写过,还是感觉挺难的,有点不会写. 这个题目大意是:给你一个数列n个元素,然后给你一天的时间,给你一个间隔时间d, 问你最少要用多少天可以把这个数列的所有元素放进去,注意元素之间必须相隔大于等于d,还有就是假设每一天之间的间隔大于d 解法: 借助队列来解题, 首先给这个元素按照元素大小来拍个序,贪心优先考虑小的,优先把小的放进去, 然后第一天肯定是要开一天的,如何后面的比这个大那就新开一天,否则就可以直接继承这一天. 这个就是解题思路,我觉得我应该…

http://acm.hdu.edu.cn/showproblem.php?pid=4803 话说C++还卡精度么?  G++  AC  C++ WA 我自己的贪心策略错了 -- 就是尽量下键,然后上键,最后下键补全,可是例子都过不了..... 题解參考http://www.cnblogs.com/xuesu/p/3967704.html http://www.cnblogs.com/Canon-CSU/p/3451784.html http://blog.csdn.net/keshuai199…

题目链接: PKU:http://poj.org/problem?id=1862 ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=543 Description Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian -…

Numbers Time Limit: 2 Seconds      Memory Limit: 65536 KB DreamGrid has a nonnegative integer n . He would like to divide n into m nonnegative integers a1,a2,...am and minimizes their bitwise or (i.e.a1+a2+...+am=n  and a1 OR a2 OR a3...OR am should…

C. Brutality time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are playing a new famous fighting game: Kortal Mombat XII. You have to perform a brutality on your opponent's character. You…

You are given a string ss consisting of exactly nn characters, and each character is either '0', '1' or '2'. Such strings are called ternary strings. Your task is to replace minimum number of characters in this string with other characters to obtain…

A few years ago Sajjad left his school and register to another one due to security reasons. Now he wishes to find Amir, one of his schoolmates and good friends. There are n schools numerated from 1 to n. One can travel between each pair of them, to d…

题目描述 最可爱的applese生日啦,他准备了许多个质量不同的蛋糕,想请一些同学来参加他的派对为他庆生,为了不让一部分同学感到不爽,他决定把每个蛋糕都分割成几份(也可以不分割),使得最小的蛋糕的质量与最大的蛋糕的质量的比值不小于一个值.但是applese的刀功并不是很好,所以他希望切尽量少的刀数使得所得到的蛋糕满足条件.由于applese为了保证每一块蛋糕的质量和期望的没有偏差,所以他一刀只能切下一块蛋糕,即将一块蛋糕分成两块,同时,他不能一刀同时切两块蛋糕,也就是说,applese一次只能将…

贪心.能凑成一组就算一组 Unrhymable Rhymes Time Limit: 10 Seconds      Memory Limit: 32768 KB      Special Judge An amateur poet Willy is going to write his first abstract poem. Since abstract art does not give much care to the meaning of the poem, Willy is plan…

L - One-Dimensional Maze Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3992 Description BaoBao is trapped in a one-dimensional maze consisting of \(n\) grids arranged in a row! The grids are num…

C - Crusaders Quest Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3983 Description Crusaders Quest is an interesting mobile game. A mysterious witch has brought great darkness to the game world,…

链接:https://ac.nowcoder.com/acm/contest/558/C来源:牛客网 小猫在研究二元组. 小猫在研究最大值. 给定N个二元组(a1,b1),(a2,b2),…,(aN,bN),请你从中选出恰好K个,使得ai的最小值与bi的最小值之和最大. 请输出ai的最小值与bi的最小值之和 输入描述: 第一行两个正整数N,K,表示二元组数量与需要选的数量. 接下来N行,第i行两个正整数ai,bi. 输出描述: 一行一个正整数,表示最大的a_i的最小值与b_i的最小值之和. 示例…

题目链接: http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=3946 题解: 用dijkstra跑单元最短路径,如果对于顶点v,存在一系列边(ui,v)使得dis[v]最小(dis[v]表示0到v的距离).这些边能且只能选一条,那么我们自然应该选cost最小的那个边了. #include<iostream> #include<cstdio> #include<cstring> #include<…

Optiver sponsored problem. After years of hard work Optiver has developed a mathematical model that allows them to predict wether or not a company will be succesful. This obviously gives them a great advantage on the stock market. In the past, Optive…