C 解法, 执行用时: 3ms, 内存消耗: 336KB, 提交时间: 2022-06-20

#include <stdio.h>

int max(int a, int b) {
    return a > b ? a : b;
}

int min(int a, int b) {
    return a > b ? b : a;
}

int main() {
    int n, k;
    scanf("%d %d", &n, &k);

    int *nums = (int *)malloc(sizeof(int) * n);
    int *dp = (int *)malloc(sizeof(int) * 2 * k);
    int *ndp = (int *)malloc(sizeof(int) * 2 * k);
    for (int i = 0; i < n; i++) {
        scanf("%d", &nums[i]);
        
        if (i == 0) {
            for (int j = 0; j < 2 * k; j+=2) {
                dp[j] = -nums[0];
            }
            for (int j = 1; j < 2 * k; j+=2) {
                dp[j] = 0;
            }
        } else {
            for (int j = 0; j < 2 * k; j+=2) {
                ndp[j] = max(dp[j], (j == 0 ? 0 : dp[j-1])-nums[i]);
            }
            for (int j = 1; j < 2 * k; j+=2) {
                ndp[j] = max(dp[j], dp[j-1] + nums[i]);
            }
            for (int j = 0; j < 2 * k; j++) {
                dp[j] = ndp[j];
            }
        }
    }
    
    int maxProfit = 0;
    for (int j = 1; j < 2 * k; j+=2) {
        if (dp[j] > maxProfit) {
            maxProfit = dp[j];
        }
    }
    
    printf("%d", maxProfit);

    free(dp);
    free(ndp);
    free(nums);
    
    return 0;
}

C 解法, 执行用时: 3ms, 内存消耗: 344KB, 提交时间: 2022-06-19

#include<stdio.h>
int max(int i,int j){
    if(i>j)
        return i;
    else
        return j;
}
int main(){
    int n,k;
    scanf("%d%d",&n,&k);//n表示天数，k表示交易次数
    int prices[n];
    int dp1[k+1];
    int dp2[k+1];
    for(int i=0;i<n;i++){
        scanf("%d",&prices[i]);
    }
    for(int i=0;i<=k;i++){
        dp1[i]=0;
        dp2[i]=0;
    }
    for(int i=0;i<n;i++){
        for(int j=1;j<=k;j++){
            if(i==0){
                dp1[j]=0;
                dp2[j]=-prices[i];
                continue;
            }
            dp1[j] = max(dp1[j],dp2[j]+prices[i]);
            dp2[j] = max(dp2[j],dp1[j-1]-prices[i]);
        }
    }
    printf("%d",dp1[k]);
    return 0;
}

C 解法, 执行用时: 3ms, 内存消耗: 352KB, 提交时间: 2022-06-23

# include <stdio.h>

int Max(int a,int b)
{
    return (a>b?a:b);
}

int maxProfit(int* prices, int pricesLen,int k ) 
{
    // write code here
    int i,j,dp1[k+1],dp2[k+1];
    //dp1[k+1]第k笔交易未持股或之前已经卖出+
    //dp2[k+1]第k笔交易持股，之前已经买入-
    for(int i=0;i<=k;i++)
    {
        dp1[i]=0;
        dp2[i]=0;
    }
    for(i=0;i<pricesLen;i++)
    {
        for(j=1;j<=k;j++)
        {
            if(i==0)
            {
                dp1[j]=0;
                dp2[j]=-prices[0];
                continue;
            }
            dp1[j]=Max(dp1[j], dp2[j]+prices[i]);
            dp2[j]=Max(dp2[j], dp1[j-1]-prices[i]);
        }
    }
    return dp1[k];
}

int main()
{
    int pricesLen,k,result,i;
    scanf("%d %d",&pricesLen,&k);
    int prices[pricesLen];
    for (i=0;i<pricesLen;i++)
    {
        scanf("%d",&prices[i]);
    }
    result=maxProfit(prices, pricesLen ,k); 
    printf("%d",result);
    return 0;
}

C++ 解法, 执行用时: 3ms, 内存消耗: 408KB, 提交时间: 2022-05-31

#include <vector>
#include <cstdio>
#include<limits.h>  //头文件包含最小值最大值
int main() {
    int n, k;
    scanf("%d %d", &n, &k);
    std::vector<int> prices(n);
    for (int i = 0; i < n; ++i) {
        scanf("%d", &prices[i]);
    }
    std::vector<std::vector<int>> dp(k + 1, std::vector<int>(2));
    dp[0][0] = 0;
    dp[0][1] = INT_MIN;
    for (int i = 1; i <= k; ++i) {
        dp[i][0] = 0;
        dp[i][1] = -prices[0];
    }
    for (int i = 0; i < n; ++i) {
        for (int j = k; j >= 1; --j) {
            // 不持有状态
            dp[j][0] = std::max(dp[j][0], dp[j][1] + prices[i]);
            // 持有状态
            dp[j][1] = std::max(dp[j][1], dp[j - 1][0] - prices[i]);
        }
    }
    printf("%d\n", dp[k][0]);
    return 0;
}

C++ 解法, 执行用时: 3ms, 内存消耗: 408KB, 提交时间: 2022-03-18

#include<iostream>
#include<vector>
using namespace std;
int main() {
    int n, k;
    cin >> n >> k;
   // vector<vector<vector<int>>> dp(n, vector<vector<int>>(k + 1, vector<int>(2,
             //                      0)));
    //vector<vector<int>> dp(k+1,vector<int>(2,0));
    vector<int> dp1(k+1,0);
    vector<int> dp2(k+1,0);
    int data[n];
    for (int i = 0; i < n; i++) {
        cin >> data[i];
    }
    for (int i = 0; i < n; i++) {
        for (int j = 1; j <= k; j++) {
            if (i == 0) {
                dp1[j] = 0;
                dp2[j] = -data[i];
                continue;
            }
            dp1[j] = max(dp1[j],dp2[j]+data[i]);
            dp2[j] = max(dp2[j],dp1[j-1]-data[i]);
            //dp[i][j][0] = max(dp[i - 1][j][0], dp[i - 1][j][1] + data[i]);
            //dp[i][j][1] = max(dp[i - 1][j][1], dp[i - 1][j - 1][0] - data[i]);
        }
    }
    cout<<dp1[k];
    return 0;
}