programing pearls

Problem descriptionlee has a string of N pearls. in the beginning, all the pearls have no color. he plans to color the pearls to make it more fascinating. he drew his ideal pattern of the string on a paper and asks for your help. In each operation, he selects some continuous pearls and all these

[Original question link] Http://acm.pku.edu.cn/JudgeOnline/problem? Id = 1260 [Topic] I want to buy several kinds of valuable pearls, but I have to pay 10 more for a pearl, and if I want to buy 100 pearls worth 1, I have to pay 110 yuan. A pearl can be filled with pearls that are more expensive than it. Therefore, it may be more cost-effective to replace

Question: Given several types of pearls and their unit prices, you can buy the same number of pearls of the same (or higher) quality with the minimum amount of money. It is required that you pay (n + 10) * P (10 * P) for the n (price P) pearls of any category. (1) The number of pearls required to be purchased is cert

Question: A variety of pearls, each purchase must add 10 to the original quantity. For example, buy 5 pearls with a unit price of 10. The cost is (5 + 10) * 10 = 150. buy 100 pearls with a unit price of 20.The cost is (100 + 10) * 20 = 2200. The total cost is 150 + 2200 = 2350. if the quality of the pearl is improved. Required 105The price of

Tags: des style blog HTTP Io color ar OS Java Simple Bipartite Graph Matching: The number of edges at each position can be the number of connected edges. Game with pearls Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/65536 K (Java/Others)Total submission (s): 122 accepted submission (s): 85 Problem descriptiontom and Jerry are playing a game with tubes and pearls. The rule of the g

Here are n kinds of pearls from low to high. Each pearl will give you the expected quantity and unit price. You need to find the original fixed number of pearls, so that the grade and price are the best. (Pearl flower money for each grade = (planned quantity + 10) * unit price) Let's look at an example: 31 101 11100 12First, the original plan to buy 100 + 1 + 1 = 102 yuan spent {(1 + 10) * 10 + (1 + 10) *

Test instructions: give n a pipe, each tube has a certain number of pearls, now Jerry began to put some pearls on the pipe, put on the number of pearls must be a multiple of k, can not put. Finally, the pipe is sorted, if I can do the first pipe above the I have a pearl, then Jerry wins, and Tom wins.Idea: The data is relatively small, so I am water, simulation p

Question: There are c kinds of pearls of different quality. If you want to buy a pearl of a certain quality, you must pay 10 more for this pearl based on the quantity you have bought, high-quality pearls can be used to replace low-quality pearls to save some cost. The minimum cost for buying all target pearls (high qua

HDOJ 5009 Paint Pearls, hdoj5009 DP + optimization, because the cost is n ^ 2, so when num × num is greater than DP [I], you can jump out .... Paint Pearls Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission (s): 1245 Accepted Submission (s): 395 Problem DescriptionLee has a string of n pearls. in the beginning, all

Hdu 1300 Pearls (dp) Question: A variety of pearls, each purchase must add 10 to the original quantity. For example, buy 5 pearls with a unit price of 10. The cost is (5 + 10) * 10 = 150. buy 100 pearls with a unit price of 20. The cost is (100 + 10) * 20 = 2200. The total cost is 150 + 2200 = 2350. if the quality of t

Paint Pearlsproblem DescriptionLee has a string of n pearls. In the beginning, all the pearls has no color. He plans to color the pearls to make it more fascinating. He drew his ideal pattern of the string on a paper and asks for your help.In each operation, he selects some continuous pearls and all these

From: http://blog.csdn.net/accelerator_/article/details/39271751 Vomit blood ac... 11668627 22:15:24 Accepted 5009 1265 Ms 1980 K 2290 B G ++ Czy Paint pearlsTime Limit: 4000/2000 MS (Java/others) memory limit: 65536/65536 K (Java/others) total submission (s): 1473 accepted submission (s): 466Problem description Lee has a string of N pearls. in the beginning, all the

Test instructions: There are different grades of pearls, the price is different, now list the need to buy pearls, to spend the least money (can buy high-grade to replace low-grade)Analysis: Dp[i] represents the minimum amount of money to spend from the lowest level to the I level, dp[i]=min (Dp[j]+v) v is the money from j+1 to i all buy I grade flowers. Requires a triple loop.Code:#include HDU 1300

value for(intK=i; k1][J] + sum[j]-sum[i-1]);//State transfer equation F[i][j] = min{f[i][k] + f[k + 1][j] + sum[j]-sum[i-1]} | I G[I][J] = max (G[i][j],g[i][k] + g[k +1][J] + sum[j]-sum[i-1]); }if(Ans_maxintMain () {scanf ("%d", n); for(intI=1; i"%d", w[i]); W[i+n]=w[i]; } for(intI=1; i2*n; i++) {Sum[i]=sum[i-1]+w[i]; } DP (); for(intI=1; iif(Ans_min>f[i][i+n-1]) Ans_min=f[i][i+n-1]; } printf ("%d\n", ans_min); printf"%d\n", Ans_max);return 0;}The above code is written in the fir

This article is an appendix of programming pearls. I have selected several practical excerpts as follows:Http://www.cs.bell-labs.com/cm/cs/pearls/apprules.html Space-for-time rules (space for Time ):Data Structure augmentation.The time required for common operations on data can be written ced by augmenting the structure with extra information or by changing the information within the structure so that it ca

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After reading programming pearls second editionSummary and application of personal practices.1. The main problem for programmers is not necessarily technical, but also psychological:He is trying to solve a wrong problem, so he cannot make progress.By breaking the conceptual barriers, we can solve a simpler problem.We finally solved the problem.2. "The more common the problem is, the easier it will be to solve it." for programming,This means to directl

By intuition Guess, wrote a DP, incredibly can AC.DP[I][J] means the minimum total cost of the item I to the nth item is purchased and the minimum unit price is the first type of article J.#include #include#include#includeusing namespacestd;Const Long Longinf=999999999999999;Const intmaxn= -+Ten;intT,n;Long LongDP[MAXN][MAXN];Long LongA[MAXN],C[MAXN];intMain () {scanf ("%d",T); while(t--) {scanf ("%d",N); for(intI=1; i"%lld%lld",a[i],C[i]); for(intI=1; i) for(intj=1; jINF; Dp[n][n

Test instructions: Jerry, Tom play games, give you out of n boxes, a[i] for the initial, the number of small balls in the first. Jerry can then add a 0 or K-fold ball to each box, and after that, Jerry will rearrange the boxes, and if I have a small ball in the box I, Jerry will win, output "Jerry" or Output "Tom".Analysis: First count each number of balls have how many boxes, then from small to Oita past, the rest of the box to i+k position, so scan the array, if more than one and more than a n

State transfer: dp[i] = min (Dp[i], dp[j] + price[i]* (sum[i]-sum[j]+10))　　　　1#include 2#include 3#include 4 using namespacestd;5 Const intN =101;6 7 intN;8 intT;9 intSum[n], dp[n], price[n];Ten One intMain () { Ascanf"%d", t); - while(t--) { -scanf"%d", n); the for(inti =0; I ) { -scanf"%d%d", sum[i], price[i]); - if(I >0) { -Sum[i] + = sum[i-1]; + } - } + A for(inti =0; I ) { atDp[i] = (Sum[i] +Ten) *Price[i]; - for(intj =0;