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Description: Say you have an array for whichITh element is the price of a given stock on dayI. If you were only permitted to complete at most one transaction (ie, buy one and every one share of the stock), design an algorithm to find the maximum profit. Solution: You only need to calculate whether the difference between the current value and the previous minimum value is greater than the maximum benefit. The following code is used: 1 class Solution {

Say you have an array for whichITh element is the price of a given stock on dayI. If you were only permitted to complete at most one transaction (ie, buy one and every one share of the stock), design an algorithm to find the maximum profit. The problem is abstracted to find the maximum difference value of the array and satisfy the condition that the values are smaller, larger, and later. First, I wrote the following code to test whether leetcode has s

)); Memset (C, 0, sizeof (INT) * (n + 1 )); TBL [0] = 1; C [0] = 1; For (INT I = 1; I { TBL [I] = 1; For (Int J = 0; j { If (TBL [J] =-1) continue; If (ARR [J]> arr [I] TBL [I] { TBL [I] = TBL [J] + 1; } } For (Int J = 0; j { If (ARR [J]> arr [I] TBL [I] = TBL [J] + 1) { C [I] + = C [J]; } } If (C [I] = 0) C [I] = 1; // when increasing /* You do not need the following code. For (Int J = 0; j { If (ARR [I] = arr [J] TBL [I] = TBL [J] C [I] = C [J]) { TBL [I] =

Poj 1952 buy low, buy lower maximum descent sub-sequence count, pojlower Question: Calculate the maximum length and number of sub-sequences for n. Analysis: Dp to avoid repeated Counting during counting. Code: // Poj 1952 // sep9 # include

This question has two questions, the first question is the longest descending subsequence.For the second ask the number of the longest descending sequence can be solved by asking the first question process. Set Maxcnt[i] to be the number of the longest descending sequence at the end of item I.For all J (1≤j≤i-1) if there is (S[j]>s[i] and maxlength[j]+1>maxlength[i]) then maxcnt[i]=maxcnt[j], otherwise if (maxlength[j]+1= = Maxlength[i]) can take advantage of the addition principle, maxcnt[i]=ma

The minimum spanning tree algorithm is simpleJust adding something new, there is some deeper understanding of the need for minimal spanning tree algorithms and the series of checks used.Some complex problems of the method#include Given the need to enumerate the sub-algorithms using several options.The workaround in the above. It uses a subset of the help of the binary counting method. A subset of the enumeration algorithm applies only to a relatively small set of elements.Taking the structure ab

CODE:#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.PAT 1092. To buy or don't to buy (STL)

;MP; for(inti=n;i>=1; i--){ if(Mp[a[i]]) next[i]=Mp[a[i]]; Elsenext[i]=-1; Mp[a[i]]=i; } REP (I,1, N) {Dp[i]=1; REP (J,1, I-1){ if(a[j]>A[i]) {Dp[i]=max (dp[i],dp[j]+1); }}} REP (I,1, N) cnt[i].init (); REP (i,1, N) { if(dp[i]==1) {Cnt[i].eq (1); Continue; } REP (J,1, I-1){ if(a[j]>A[i]) { if(dp[j]+1==Dp[i]) { if(next[j]!=-1next[j]Continue; Cnt[i]=Add (Cnt[i],cnt[j]); } } }

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Say you have an array for whichITh element is the price of a given stock on dayI. If you were only permitted to complete at most one transaction (ie, buy one and every one share of the stock), design an algorithm to find the maximum profit. Idea: traverse the input array in sequence to record the current minimum value. The maximum value of the difference between each element and the current minimum value is required. 1 class solution {2 public: 3 i

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Question address: http://poj.org/problem? Id = 1952 Buy low, Buy lower Time limit:1000 ms Memory limit:30000 K Total submissions:6692 Accepted:2302 Description The advice to "buy low" is half the formula to success in the bovine stock market. To be considered a great investor you must also follow this probl

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Buy low, buy LOWER Time Limit: 1000MS Memory Limit: 30000K DescriptionThe advice to ' buy low ' is half the formula-success in the ' bovine stock market. To is considered a great investor you must also follow this problems ' advice: "Buy