2024-07-12
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Link to the respondentem quaestionem:188. Optimum tempus emere et vendere nervo IV - LeetCode
Niuke nexus thematis respondentis:Optimum tempus emere et vendere nervo (4)_Niuke Topic_Niuke.com (nowcoder.com)
Ut clarius distinguamusemptumetvendere, repone cumHave stocksetnulla stirpeDuae civitates.
- f[i][j]: Nequaquam. ego In fine diei factum j transaction, currently inHave stocksMaximum commodum rei publicae.
- g[i][j]: Nequaquam. ego In fine diei factum j transaction, currently innulla stirpeMaximum commodum rei publicae.
for* f*[i][j]sunt etiam duae condiciones in quibus ego Post dies completus j Res, in manu ad hoc tempusHave stocksstatus:
- exist i-1 in caelo, in manibusHave stockset negotiatus est j SECUNDUS. Pridie i-mo die nihil.Reditus in hoc tempore est f[i - 1][j]。
- exist i-1 in caelo, in manibusnon stockset negotiatus est j SECUNDUS.In primisego Cum sol cecidit, uncias emi tessera.Sicego In fine diei ligna funt.Reditus in hoc tempore estg[i - 1][j] - prices[i]。
In his duobus casibus quid opus est?maximum valoremita status aequationis ipsius f transitus est;
f[i][j] = max(f[i - 1][j], g[i - 1][j] - prices[i]
for* g*[i][j], duas sequentes condiciones in quibus possumus ego Post dies completus j Res, in manu ad hoc tempusnon stocksstatus:
- exist i-1 In die diei lignum in manibus meis non habui et eas dedi. j SECUNDUS.I-th die nihil .Reditus in hoc tempore estg[i - 1][j]。
- exist i-1 Die aurorae nervum habui in manibus meis, et dedi ea. j - 1 SECUNDUS.In primisego Cum serenum est, put Truncus venditus.Sicego In fine diei, negotiamur j SECUNDUS.Reditus in hoc tempore est f[i - 1][j - 1] + pretia[i]。
In his duobus casibus quid opus est?maximum valorem,ergo g* Aequatio status transitus est:
g[i][j] = max(g[i - 1][j], f[i - 1][j - 1] + prices[i])
Negotiatio inter eos talis est:
Imple in utroque ordine a summo ad imum, ordinem quemque a sinistro ad dextrum, et imple utrasque tabulas simul.
Redit valorem maximum in re publica venditionis, sed quotiens negotiatus est nescimus, unde redit g* Maxima vis in ultimo ordine tabulae.
Numerus negotiorum noster dimidium totius numeri dierum non excedens, primum ergo possumus k Cum ea agamus et magnitudinem problematum optimize:k = min(k, n / 2)。
- //力扣
- //【动态规划-二维dp-2个状态】
- class Solution {
- //f[i][j]:第i天结束后,完成了j次交易,此时处于“买入”状态下的最大利润
- //g[i][j]:第i天结束后,完成了j次交易,此时处于“卖出”状态下的最大利润
- private:
- const int INF=0x3f3f3f3f;
- public:
- int maxProfit(int k, vector<int>& prices) {
- int n=prices.size();
- k=min(k, n/2); //优化:处理最多交易次数
- vector<vector<int>> f(n, vector<int>(k+1, -INF));
- vector<vector<int>> g(n, vector<int>(k+1, -INF));
- f[0][0]=-prices[0], g[0][0]=0;
- for(int i=1; i<n; i++)
- {
- for(int j=0; j<=k; j++)
- {
- f[i][j]=max(f[i-1][j], g[i-1][j]-prices[i]);
- g[i][j]=g[i-1][j];
- if(j>=1) g[i][j]=max(g[i][j], f[i-1][j-1]+prices[i]);
- }
- }
- int res=0;
- for(int j=0; j<=k; j++)
- res=max(res, g[n-1][j]);
- return res;
- }
- };
-
- //【动态规划-二维dp-2k+1个状态】
- class Solution {
- //dp[i][0] -- 没有操作
- //下面j为奇数:买入;j为偶数:卖出 (j的范围:1~2k-1)
- //dp[i][j] -- 第1~k次买入
- //dp[i][j+1] -- 第1~k次卖出
- public:
- int maxProfit(int k, vector<int>& prices) {
- int n=prices.size();
- vector<vector<int>> dp(n, vector<int>(2*k+1));
- for(int j=1; j<2*k; j+=2)
- dp[0][j]=-prices[0];
- for(int i=1; i<n; i++)
- {
- for(int j=0; j<2*k; j+=2)
- {
- dp[i][j+1]=max(dp[i-1][j+1], dp[i-1][j]-prices[i]);
- dp[i][j+2]=max(dp[i-1][j+2], dp[i-1][j+1]+prices[i]);
- }
- }
- return dp[n-1][2*k];
- }
- };
- //牛客
- #include <iostream>
- #include <cstring>
- using namespace std;
-
- const int INF=0x3f3f3f3f;
- const int N=1010, M=110;
- int prices[N];
- int f[N][M], g[N][M];
- //f[i][j]:第i天结束后,完成了j次交易,此时处于“买入”状态下的最大利润
- //g[i][j]:第i天结束后,完成了j次交易,此时处于“卖出”状态下的最大利润
-
- int main()
- {
- int n, k;
- cin >> n >> k;
- for(int i=0; i<n; i++)
- cin >> prices[i];
- memset(f, -INF, sizeof(f));
- memset(g, -INF, sizeof(g));
- int res=0;
- f[0][0]=-prices[0], g[0][0]=0;
- for(int i=1; i<n; i++)
- {
- for(int j=0; j<=k; j++)
- {
- f[i][j]=max(f[i-1][j], g[i-1][j]-prices[i]);
- g[i][j]=g[i-1][j];
- if(j>0) g[i][j]=max(g[i][j], f[i-1][j-1]+prices[i]);
- res=max(res, g[i][j]);
- }
- }
- cout << res << endl;
- return 0;
- }
-
- //值得学习的代码
- #include <iostream>
- using namespace std;
-
- const int N = 1010, M = 110;
-
- int n, k, p[N];
- int f[N][M], g[N][M];
-
- int main()
- {
- cin >> n >> k;
- for(int i = 0; i < n; i++) cin >> p[i];
-
- k = min(k, n / 2);
- for(int j = 0; j <= k; j++) f[0][j] = g[0][j] = -0x3f3f3f3f;
- f[0][0] = -p[0], g[0][0] = 0;
-
- for(int i = 1; i < n; i++)
- {
- for(int j = 0; j <= k; j++)
- {
- f[i][j] = max(f[i - 1][j], g[i - 1][j] - p[i]);
- g[i][j] = g[i - 1][j];
- if(j >= 1) g[i][j] = max(g[i][j], f[i - 1][j - 1] + p[i]);
- }
- }
-
- int ret = 0;
- for(int j = 0; j <= k; j++) ret = max(ret, g[n - 1][j]);
-
- cout << ret << endl;
-
- return 0;
- }
Series quaestionum quasi nervum emptionis et venditionis accuratam familiaritatem cum punctorum cognitionis requirit.