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Link:https://uva.onlinejudge.org/index.php?option=com_onlinejudgeItemid=8page=show_problemproblem= 1995Test instructionsIn a straight line there are N (2≤n≤100000) equidistant villages, each of which buys wine or sells liquor.The demand for wine in the first village is Ai ( -1000≤ai≤1000), of which ai>0 means to buy wine and aiTransporting K-units of wine from one village to the neighboring village requires a workforce of k units.Calculating at least how much labor is needed to meet the needs of

. Sample Input 31+2*3*4+54*18+14+7*103+11+4*1*13*12*8+3*3+8 Sample output The maximum and minimum are 81 and 30.The maximum and minimum are 1560 and 156.The maximum and minimum are 339768 and 5023. Question: Add parentheses to an expression that contains only the plus (+) and minus (*) operators. Then, add parentheses to obtain the maximum and minimum values of the expression. Analysis and Summary: If the maximum value is required, the priority of + is greater than *. If the minimum value is

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Say you has an array for which the i-th element is the price of a given-stock on day I.If you were-permitted-to-complete at most one transaction (ie, buy one and sell one share of the stock), design an AL Gorithm to find the maximum profit.Simple dynamic programming problem, but at first I didn't read the meaning of the topic. The buy here should of course be before the sale, so that is not the kind to take a max a min subtraction can be solved, the code is as follows:1 classSolution {2 Public:

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Test instructions: Given a sequence, the element I represents the price of this stock for the first day, asking how much money can be earned at the best time to buy and sell? Buy only once, and only 1 shares, assuming unlimited cost.Idea: To find a low-priced time to buy, at the highest price when the sale of profits will be the biggest. But time is not a conflict, for example buy tomorrow, sell today. Therefore, for today's price, we should find the lowest price of the stock before today, buy,

#include 3 4 Long LongScan () {//input Plug -in5 intres =0, flag =0;6 Charch;7 if(ch = getchar ()) = ='-') flag =1;8 Else if(Ch >='0' CH '9') res = CH-'0';9 while(ch = getchar ()) >='0' CH '9')Tenres = res *Ten+ (CH-'0'); One returnFlag? -Res:res; A } - - intMain () the { - //freopen ("In.txt", "R", stdin); - - intN; + while(SCANF ("%d", n) = =1N) - { + GetChar (); A Long LongAns =0, last =0; at intA; - for(inti =0; I i) -

Title: 11054-wine Trading in GergoviaThe owner of each house either buys the wine or sells the wine, and the number of the wines they buy and sell is the same. Liquor-making people need to transport wine to the home of the person who buys the wine, and every bottle of wine transported to the adjoining house costs a unit of money, asking how to make the overall shipping cost the least.Problem-solving ideas: Both sell and buy, need to be transported. Th

array every time (the element that starts with this array must be the coordinates of the leftmost person who buys the wine), if the item in front of the array is 0, we move the start coordinate to the right (without this step, it will definitely time out).but just now on the internet to see another one way to do this is not to consider the purchase and sale relationship stipulated in the topic, and to stipulate that the first person buys the wine of the first i+1 person, so the answer is correc

= Prompt.getmarginmethod (); return Marginmethod = = null? Null:marginmethod; } @Transient Public Feemethod Getmarginfeemethod () {Feemethod marginfeemethod = null; if (prompt! = null) Marginfeemethod = Prompt.getmarginfeemethod (); return Marginfeemethod = = null? Null:marginfeemethod; } @Transient protected Double getpricebasis () {double pricebasis = 0; if (prompt! = null) Pricebasis = Prompt.getpricebasis (); return pricebasis = = 0? GetQuote (). Getbasi

Monotone Queue Optimization DP template problemIt is not difficult to list the DP equation:For the case of buyingDue to dp[i][j]=max{dp[i-w-1][k]+k*ap[i]-j*ap[i]}Ap[i]*j is fixed, the monotony of maintaining dp[i-w-1][k]+k*ap[i in the queue can be1#include 2#include string.h>3#include 4 using namespacestd;5 Const intMAXN = .;6 intt,maxp,w,ap[maxn],bp[maxn],as[maxn],bs[maxn],dp[maxn][maxn],q[maxn*Ten]; 7 8 intMain () {9scanf"%d%d%d", t, AMP;MAXP, W);Ten for(intI=1; i) Onescanf"%d%d%d%d", ap[

Question connection: 10700-camel trading A formula containing only '+' and '*' is provided. The maximum and minimum values of the formula must be obtained without considering the priority. Solution: the maximum value is nothing more than processing all addition. The minimum value is used to process all multiplication (because all values are positive numbers ). # include # include using namespace STD; int main () {int

variables2. Configure Python script environment variables3. Download the Setuptools, unzip to any directory4. Go to the directory above5.Python setup.py Install6. Verify Easy_install–showhelp7.easy_install Common CommandsEasy_install installation Package-U update-M Delete Library8.PIP installation is similar to Easy_installPIP Install package9. If an ASCII-related exception occursTo modify the lib\mimetype.py file under the Python installation path, add the following lines under import:

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For you to trade cat software users to detailed analysis to share the meaning of the voucher. Analytical sharing: A voucher is a marketing tool for dealing with a cat's platform, a trading partner with a voucher, when purchasing a specified product, the corresponding denomination of the voucher can be used to deduct cash payments. Voucher can be used in the validity period, can check my voucher in the personal center. The buyer used a voucher

For you to trade cat software users to detailed analysis of the time required to share the account. Analytical sharing: The trading cat is in deep cooperation with Alipay and your bankroll management is completed by Alipay assistance. At present, Alipay has 3 kinds of presentation methods: Ordinary, 2 hours to bring the present, real-time. General mention: The average person is 2-3 days to account; the company is 1-2 days. 2 hours to bring:

1 into the mobile phone's Ocean Dock interface Click on the bottom of the "my" entry and then click "Set Trading Password" and then open to enter. (pictured below) 2 in the pop-up password we enter and then click "Next", will send the verification code to the phone, fill in the Sent verification code click "Next". (pictured below) 3 Enter the same password two times and then click "Confirm". (pictured below) Good to here about this transacti

the trading plan of the bear Time limit:12000/6000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 374 Accepted Submission (s): 122 Problem Description the bear was involved in the business meeting of the meow-haha village, but the business conference encountered a dilemma: Meow haha village and the surrounding villages can be seen as a total of n pieces of area, M highway composed of areas. Due to the difference

The app/cli sends a INSTANTIATE/INVOKE request to the specified endorser node, and after endorser receives the request, if the relevant chaincode exists, the request is sent to The Chaincode side, and executes the correlation function (because the execution process, may involve multiple state reads and writes, and each read and write will involve the operation of the underlying DB, so this process involves multiple communication with the endorser node); After Chaincode execution is complete, Sen

Title Description This is the American Regiment's 2016 recruitment of pen questions. Title Description In the trading day of the stock market, it is assumed that a maximum of two transactions can be made (that is, the number of purchases and sales is less than or equal to 2), and the rule is that a deal must be executed after another (i.e. buy-sell-buy-sell order). Given the stock change sequence of the day, write a program that calculates the maxim