Dynamic Programming - stock trading

Stock trading

Meaning of the questions: tell the N-day stock prices, were two transactions, multiple operations can be performed on the same day, but then you must first buy, you must first sell before they can buy a second time. Q. What is the maximum profit?

1. dp [i] i days before the first transaction to maximize profits
• dp[i] = max(dp[i-1],num[i]-minn)
• Is not the first day i sell: i-1 equal before the day for the first time get the maximum profit exchange
• I was the first day sold: maximum profit for minimum price minus the n-th day before the stock price of n days
2. DP2 is [i] for the i-th day to obtain the maximum value of the sum of the two exchanges the second profit
• dp2[i]=max(dp[i],dp2[i-1]+num[i]-num[i-1])
• Was the second day i buy: it is the sum of two days ago i once the largest trading profit, trading profit for the second time 0
• Is not the first day i bought: i-1 equal to the first day to sell, plus the difference between the stock price and the i-th day of the first i-1 stock price of the day
3. If the transaction only once a day, similar to the solution. dp2 [i] is the second consideration is not the first i-1 days to buy.   
• dp2[i] =max(dp[i-2]+num[i]-num[i-1],dp2[i-1]+num[i]-num[i-1])

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <string.h>
#include <assert.h>
#include <queue>
#include <vector>
#include<list>
#include <algorithm>
#include<iostream>
#include<math.h>
using namespace std;
#define N 100005
#define Inf 0x3f3f3f3f
int num[N];
int dp[N];
int dp2[N];
int minValue[N];

int main() {
  int t;
  cin>>t;
  while(t--){
      int n;
      scanf("%d",&n);;
      memset(num,0,sizeof(num));
      memset(dp,0,sizeof(dp));
      memset(dp2,0,sizeof(dp2));
      memset(minValue,0,sizeof(minValue));
      int maxx = -1;
      int minn =Inf;
      for(int i=1;i<=n;i++){
          scanf("%d",&num[i]);
          if(minn>num[i]){
              minn = num[i];
          }
        dp[i] = max(dp[i-1],num[i]-minn);
      }

    for(int i=2;i<=n;i++){
        dp2[i] = max(dp[i],dp2[i-1]+num[i]-num[i-1]);
        if(maxx<dp2[i]){
            maxx= dp2[i];
        }
    }
  
  cout<<maxx<<endl;
  
      
  }
  return 0;
}