/ 卡路里

Problem statement:

问题陈述：

Shivang is very foodie but he has a diet plan. He has an array of elements indicating the calorie of food he can consume on that day. In his diet plan, he can’t eat on for three consecutive days. But since Shivang is very foodie, so he decided to take as much as calorie while remembering the fact that he cannot eat on three consecutive days. Help him in finding the number of calories he could take.

Shivang是位美食家，但他有饮食计划。 他具有一系列元素，这些元素指示他当天可以吃的食物的热量。 在他的饮食计划中，他连续三天不能进食。 但是由于Shivang是一位非常美食的人，因此他决定摄取尽可能多的卡路里，同时记住自己连续三天不能进食的事实。 帮助他找到可以摄取的卡路里数量。

Input:

输入：

n i.e number of days. In the next line is n space-separated values of a_i indicating the intake of calorie per day.

n即天数。 在下一行中， _i的 n个以空格分隔的值表示每天的卡路里摄入量。

Output:

输出：

Print the maximum amount of calorie, Shivang can take in a new line.

打印最大卡路里，Shivang可以换行。

Constraints:

限制条件：

1<=n<=100000
1<=ai<=100000

Example:

例：

Input:
2
n=5
Array input, calorie for each day:
3 2 1 10 3
Output:
18
Explanation:
Shivang will eat on 1st, 2nd, 4th and 5th day 
So, total he intakes 18 which is maximum
Input:
8
3 2 3 2 3 5 1 3 
Output:
17
Explanation:
Shivang will eat on 1st, 3rd, 5th , 6th and 8th 
day which totals (3+3+3+5+3) = 17

Solution Approach:

解决方法：

Let say,

说，

f(n) = maximum calorie can be taken till nth day under the constraint

No, let's think of the case,

不，让我们考虑一下情况，

1. If Shivang takes food on i^th day, he need to skip on (i-1)_^th day and sum with f(i-2)

   如果Shivang在^第 i ^个一天需要吃的，他需要跳过(I-1) _^个天，和与F(I-2)

2. Tale food on i^th day, (i-1)^th day and skip on (i-2)^th day, sum up with f(i-3)

   i上故事食品^个天， 第(i-1) ^个天，并跳过对第(i-2) ^个天，总结和f(I-3)

3. Skip i^th day, so take f(i-1)

   跳过^第 i ^个天，所以采取F(I-1)

So,

所以，

f(i)=maximum(f(i-1),f(i-2)+arr[i],arr[i]+arr[i-1]+f(i-3) i>=3
For base case,
f(0)=arr[0]
f(1)=arr[0]+arr[1]
f(2)=maximum(arr[0]+arr[1],arr[1]+arr[2],arr[2]+arr[0])

Converting to Dynamic programming

转换为动态编程

Now, we need to convert the above recursion into dynamic programming to avoid the overlapping sub-problem re-computation

现在，我们需要将上述递归转换为动态编程，以避免重叠的子问题重新计算

1)  Declare int DP[n];
2)  Fill with the base case
DP[0]=arr[0];
DP[1]=arr[0]+arr[1];
DP[2]=std::max(arr[1]+arr[2],std::max(arr[0]+arr[2],DP[1]));
3)  Compute the other values
for i=3 to n-1
DP[i]=maximum(arr[i]+arr[i-1]+DP[i-3],arr[i]+DP[i-2],DP[i-1]);
end for
4)  Return the result, DP[n-1];

C++ Implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
int maximumCalorie(vector<int> arr, int n)
{
int DP[n];
// base case
DP[0] = arr[0];
DP[1] = arr[0] + arr[1];
DP[2] = std::max(arr[1] + arr[2], std::max(arr[0] + arr[2], DP[1]));
for (int i = 3; i < n; i++) {
DP[i] = std::max(arr[i] + arr[i - 1] + DP[i - 3], std::max(arr[i] + DP[i - 2], DP[i - 1]));
}
return DP[n - 1];
}
int main()
{
int t, n, item;
cout << "Enter number of days\n";
cin >> n;
cout << "Enter calories\n";
vector<int> a;
for (int j = 0; j < n; j++) {
scanf("%d", &item);
a.push_back(item);
}
cout << "Maximum calorie sudhir can take: " << maximumCalorie(a, n) << endl;
return 0;
}

Output:

输出：

Enter number of days:
8 
Enter calories:
3 2 3 2 3 5 1 3 
Maximum calorie sudhir can take: 17

翻译自: https://www.includehelp.com/icp/maximum-calorie.aspx

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