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=Bottletop Surplus_bottle=Bottleif __name__=='__main__': Money= 10count, Surplus_bottletop, Surplus_bottle= MONEY//2, MONEY//2, MONEY//2Print('A total of%d yuan, this time drink%d bottles, the remaining bottle%d, the remaining bottles%d'%(Money,count,count,count)) Drinkandsurplus (Surplus_bottletop, Surplus_bottle)Print('A total of%d bottles, the remaining caps%d, the remaining bottles%d'% (Count,surplus_bo

0x3f3f3f3ftypedefLong Longll; typedef unsignedLong Longull;using namespacestd;Const intMAXN = 3e5+5; Priority_queueA;intHAS[MAXN];intMain () {Ios::sync_with_stdio (false); Cin.tie (0); Cout.tie (0); intI, J, K, M, N; CIN>>N; ll ans=0; for(intI=1; ii) {cin>>K; A.push (MP (K,i)); if(A.top (). first>=k)Continue; Ans+=k-A.top (). First; if(!Has[a.top (). Second]) {A.pop (); Has[i]=1; } Else{has[a.top (). Second]=0; Has[i]=1; }} coutEndl; return 0; } Codeforces Round #43

/*** The ticket will appear to sell more than 100 tickets, the num static can solve the problem*/ Private Static voidtest1 () {System.out.println ("Start selling Tickets" +Thread.CurrentThread (). GetName ()); Ticket Ticket=NewTicket (); Thread T1=NewThread (ticket); Thread T2=NewThread (ticket); Thread T3=NewThread (ticket); Thread T4=NewThread (ticket); T1.start (); T2.start (); T3.start (); T4.start (); System.out.println ("End of ticket sale" +Thread.CurrentThr

Tags: i++ ras span Art else add div start sumTest instructions: Some stock prices, we can choose to buy and sell, but only one operation a day, ask the maximum profit Idea: For the day, if the sale before he was smaller, we must be looking for a minimum day of buy, but do not know whether the sale is the most profitable, so we can use Multiset, this and set similar, but can store the same number, and sort

second sale of the selling point and selling points are likely to be maxind.Realize:Case1 is easy, is to run two times the best trading, merit to join. If there is no valid trade, it is recorded as 0.The best Buy and sell before Minind is recorded as PARTGAP2 after Partgap1,maxind.Case1result = Maxgap + Max (PARTGAP1, PARTGAP2);Case2 more trouble, let us explain in detail:Simple analysis reveals:The value of two transactions is Where Minleft is the m

[Plain] [Plain]/* Start the comments in the program header (to avoid any problems encountered during the submission of blog posts, the slash used to indicate that the comments have been deleted)* Copyright and version Declaration of the program* All rights reserved.* File name: txt. c* Author: liuyongshui* Question: Now there is a house in Beijing with a price of 2 million. If the house price increases by 10% every year, a software engineer can earn 0

LeetCode | Best time to buy and keep stock 2, leetcode= Question: Say you have an array for which the ITh element is the price of a given stock on day I. Design an algorithm to find the maximum profit. you may complete as your transactions as you like (ie, buy one and every one share of the stock multiple times ). however, you may not engage in multiple tran

Reprint Please specify source: http://blog.csdn.net/li396864285/article/details/51519786 "Original" 10 Yuan to buy beer 2 empty bottles for a bottle of 4 bottles for a bottle, the classic algorithm original title: 10 Yuan to buy beer, 2 empty bottles can be exchanged for a bottle, 4 caps can be exchanged for a bottle,

Console.WriteLine ("Toothpaste 5 yuan, toothbrush 2 yuan, soap 3 yuan, 100 yuan to buy these three kind of happen to spend light"); intA1 =0, B1 =0, C1 =0, sum =0; for(intA =0; A -; a++) {A1=5*A; for(intb =0; b -; b++) {B1=2*b; for(intc =0; C -; C++) {C1=3*C; if(A1 + B1 + C1 = = -) {sum+=1; Console.WriteLine ("Buy