moschino backpack

And that we know the weights of all coins of a given currency. then there is some minimum amount of money in the piggy-bank that we can guarantee. your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. we need your help. no more prematurely broken pigs! Inputthe input consists of T test cases. the number of them (t) is given on the first line of the input file. each test case begins with a line containing two integers E and F. they indicate the

-dimensional bag problem of sample Output0-11 refers to: for each item, there are two different kinds of charges; the choice of this item must be paid at the same time; for each generation Each price has a maximum value (Backpack capacity) that can be paid ). Ask how to select an item to maximize the value. Set these two prices to price 1 and price 2 respectively.The two costs required for item I are a [I] and B [I]. The maximum value (two types of

POJ--2063 -- investors -- backpack, 002063 shares Investment Time Limit:1000 MS Memory Limit:30000 K Total Submissions:8733 Accepted:2984 DescriptionJohn never knew he had a grand-uncle, until he got the notary's letter. he learned that his late grand-uncle had gathered a lot of money, somewhere in South-America, and that John was the only inheritor. John did not need that much money for the moment. but he realiz

ZOJ3524: Crazy Shopping (Topology Sorting + full backpack) and zoj3524shopping Because of the 90th anniversary ofCoherent Cute Patchouli(C. C.P ),Kawashiro NitoriDecides to buy a lot of rare things to celebrate. Kawashiro NitoriIs a very shyKappa(A type of water sprite that live in rivers) and she lives onYoukai Mountain.Youkai MountainIs a dangerous place fullYoukai, So normally humans are unable to be close to the mountain. But because of the fin

the Piggy-bank are 100.This is impossible.#include #include#includeusing namespacestd;Const intinf=0x3f3f3f3f;Const intmaxn=10005;intn,w;intV[MAXN],W[MAXN];intDP[MAXN];intMain () {intT; scanf ("%d",T); while(t--) { inte,f; scanf ("%d%d",e,f); scanf ("%d",N); for(intI=0; i) {scanf ("%d%d",v[i],W[i]); } W=f-e; for(intI=0; i) {Dp[i]=inf;//Prerequisite: The backpack is full. Because the value of the request is minimal, you need to initialize

01 backpack. Start your backpack JourneyThe normal backpack isI 1....NJ m ..... 0Dp[i][j]=max (Dp[i-1][j],dp[i-1][j-w[i]]+v[i]);But this range is bigger.If you use two-dimensional hyper-memorycan be transformed into one-dimensionalDp[j]=max (Dp[j],dp[j-w[i]]+v[i]);#include #includestring.h>#include#include#include#includeusing namespacestd;#defineN 350

There is a tree with a number of N, and the edge of the tree is right. Give you a positive integer k within 0~ N, you want to select K points in this tree, dye it black, and turn the other n-k Dianran into white. After all points are dyed, you will get the distance between the black dots 22 plus the white Point 22 distance and the benefit. Ask what the maximum benefit is.SolutionCompare the classic tree knapsack problem.If only the points are analyzed, the situation becomes very cumbersome and d

http://acm.hdu.edu.cn/showproblem.php?pid=2639The main idea is, to the backpack race bones, for the K-optimal solution, on the basis of the ordinary 01 backpack, add one-dimensional space, then f[i,v,k] can be understood as the first I items, put in the capacity v backpack, the value of the K-optimal solution. The time complexity is O (NVK).Talk is cheap.Look at

In order to save the lives of compatriots in the disaster area, you are prepared to purchase some food support disaster areas, now suppose you have a total of funds N, and the market has m rice, each rice is bagged products, its price range, and can only buy the whole bag.Excuse me: How many kilograms of grain can you purchase with limited funds?Sample INPUT1 8 2 2 100 4 4 100 2Sample Output4001#include 2#include 3#include string.h>4#include 5 using namespacestd;6 Const intmaxn= the;7 8 intDP[MA

world;Thanks to the opponents, they make us keep making progress and efforts.Likewise, we would like to thank the pain and hardship brought to our wealth ~ Input data first contains a positive integer c, which indicates that there is a C set of test cases, the first line of each set of test cases is two integers n and m (1Output for each set of test data, export the maximum weight that can be purchased for rice, and you can assume that you are spending more than all of the rice, and you c

Sumsets Time Limit: 2000MS Memory Limit: 200000K Total Submissions: 15045 Accepted: 5997 DescriptionFarmer John commanded his cows to search for different sets of numbers the sum to a given number. The cows use is numbers that is an integer power of 2. Here is the possible sets of numbers that sum to 7:1) 1+1+1+1+1+1+1 2) 1+1+1+1+1+2 3) 1+1+1+2+2 4) 1+1+1+4 5) 1+2+2+2 6) 1+2+4 Help FJ count all possible representations for a given i

There is a backpack can be put into the item weight of S, existing n items, the weight is w1,w2,w3,...wn. Ask if you can select several items from the N-item into the backpack, so that the sum of the weights placed is just s. If there is a choice to meet the conditions, then the backpack has a solution, or the knapsack problem is not solved.DescriptionFirst act i

Title: Uva-10280old Wine into New bottles (full backpack + pruning)The main topic: Now to put the old wine into the new bottle, each bottle has the smallest maximum capacity requirements, and then give you l wine, in the give you n bottles, each bottle of the specification also give you, each kind of bottle supply is unlimited, ask how to arrange these wine to make the remaining wine the least.Problem-solving ideas: The problem is a complete

This is a simple 01 backpack, but the knapsack problem is fairly simple, is the front of the details of the time to pay attention to, test instructions roughly said three restrictions bar1. Only A, B, c three types of invoices can be reimbursed, the other will not be reimbursed2. No more than 600 reimbursement for item items3. The total amount of each invoice does not exceed 1000With these three, there is one to be careful is the amount of reimburseme

-negative integerV P q(wherevindicates the price of the item (v),Pindicates the importance of the item ( 1~5 ),QIndicates whether the item is a main piece or an attachment. Ifq=0, indicating that the item is the main piece, ifq>0, indicating that the item is an attachment,Qis the number of the owning main part)outputs description output Description there is only one positive integer, which is the maximum value of the sum of the product of the price and the importance of the item not exceeding th

, respectively the smartness and funness for each cow.Output* Line 1:one integer:the optimal sum of TS and TF such that both TS and TF is non-negative. If no subset of the cows has non-negative TS and non-negative TF, print 0.Sample Input5-5 78-66-32 1-8-5Sample Output8HintOUTPUT DETAILS:Bessie chooses cows 1, 3, and 4, giving values of TS = -5+6+2 = 3 and TF= 7-3+1 = 5, so 3+5 = 8. Note that adding cow 2 would improve the valueof TS+TF to ten, but the new value of TF would was negative, so it's

View code /* Give you n items, each of which has a certain volume, give you a certain total volume of a certain backpack, ask you to use this backpack to install these items, there are several different combinations of items in a backpack. One combination is legal: The remaining backpack capacity must be the mini

Given n items with size a[i] and value v[i], and A backpack with size M. What's the maximum value can you put into the backpack?NoteYou cannot divide item in small pieces and the total size of items you choose should smaller or equal to M.ExampleGiven 4 items with size [2, 3, 5, 7] and value [1, 5, 2, 4], and a backpack with size 10.The maximum value is 9. Soluti

Test Instructions: The value of 1,2,3,4,5,6 items respectively a[1],a[2],a[3],a[4],a[5],a[6], ask can be divided into two piles of equal value. Solution: Convert to multiple backpacks1#include 2#include string.h>3#include 4#include 5 using namespacestd;6 7 intdp[120010];8 inta[Ten];9 Ten intNvalue; One A //0-1 backpack, at cost, to get the value of weight - voidZeroonepack (intCostintweight) - { the for(inti=nvalue;i>=cost;i--) -Dp[i]=max (dp[i]

/* Poj 1276 the meaning of multiple knapsack questions is probably given a currency m to be raised, given the number of currencies owned N [I] and the value V [I], q: How much can you raise is a multi-backpack problem DP [J] indicates the maximum number of valuable items that a backpack with a capacity of J can hold. Generally, the solution to multiple backpacks is to convert them. complete