chatGPT 刷新题完全不行啊

把题目贴出来，我用我的 chatgpt plus 跑一下

所以说离通用智能还是很远的

如果你用的是 3.5 ，那很正常，如果是 4.0 ，可能要改一下题目，虽然题目是新题，但是其实算法都是那些东西。另外让他跑测试用例，就有点扯了。。

请出题

@liqinliqin
@wonderfulcxm

2355. Maximum Number of Books You Can Take


You are given a 0-indexed integer array books of length n where books[i] denotes the number of books on the ith shelf of a bookshelf.

You are going to take books from a contiguous section of the bookshelf spanning from l to r where 0 <= l <= r < n. For each index i in the range l <= i < r, you must take strictly fewer books from shelf i than shelf i + 1.

Return the maximum number of books you can take from the bookshelf.



Example 1:

Input: books = [8,5,2,7,9]
Output: 19
Explanation:
- Take 1 book from shelf 1.
- Take 2 books from shelf 2.
- Take 7 books from shelf 3.
- Take 9 books from shelf 4.
You have taken 19 books, so return 19.
It can be proven that 19 is the maximum number of books you can take.
Example 2:

Input: books = [7,0,3,4,5]
Output: 12
Explanation:
- Take 3 books from shelf 2.
- Take 4 books from shelf 3.
- Take 5 books from shelf 4.
You have taken 12 books so return 12.
It can be proven that 12 is the maximum number of books you can take.
Example 3:

Input: books = [8,2,3,7,3,4,0,1,4,3]
Output: 13
Explanation:
- Take 1 book from shelf 0.
- Take 2 books from shelf 1.
- Take 3 books from shelf 2.
- Take 7 books from shelf 3.
You have taken 13 books so return 13.
It can be proven that 13 is the maximum number of books you can take.

废话生成器而已，不要抱有幻想

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def max_books(books):
n = len(books)
max_books = max(books)
# dp[i][j] = max books we can take if we take j books from shelf i
dp = [[0]*(max_books+1) for _ in range(n+1)]
for i in range(1, n+1):
for j in range(1, max_books+1):
if j > books[i-1]: continue
for k in range(j):
dp[i][j] = max(dp[i][j], dp[i-1][k] + j)
return max(max(row) for row in dp)

books = [8,5,2,7,9]
# books = [7,0,3,4,5]
# books = [8,2,3,7,3,4,0,1,4,3]
print(max_books(books)) # Output: 19


不知道解法正确不，但是我测试这几个例子输出一致，切特急皮特 给的答案

@Alan3 #8 上一条好像格式乱了： https://imgur.com/a/ZGJgmaz

@Alan3 #9 [img][/img]

吃饭的时候，大致看了一下 chatgpt 给出的代码，解法清晰，嘎嘎正确！！

你好，这是一个有趣的问题。根据我从网上搜索的信息，这个问题可以用动态规划来解决 https://www.luogu.com.cn/problem/CF1433B https://songhayoung.github.io/2022/08/02/PS/LeetCode/maximum-number-of-books-you-can-take/。动态规划的思路是，用一个数组 dp[i]表示从第 0 本书到第 i 本书能取走的最大数量，那么 dp[i]可以由 dp[j]和 books[j+1...i]之间的关系得到，其中 j 是满足 books[j+1...i]是一个递增序列的最小下标。具体地，如果 books[j+1...i]是一个递增序列，那么 dp[i] = dp[j] + books[i] - books[j+1] + 1 ，否则 dp[i] = dp[i-1]。最后，返回 dp[n-1]即可。

下面是一个用 Python 写的可能的代码实现：

def max_books(books):
n = len(books) # the number of books
dp = [0] * n # the array of maximum books
for i in range(n):
j = i - 1 # the index of the previous book
while j >= 0 and books[j] >= books[j+1]: # find the smallest j such that books[j+1...i] is increasing
j -= 1
if j >= 0: # if such j exists
dp[i] = dp[j] + books[i] - books[j+1] + 1 # update dp[i] by adding the difference of books and one
else: # otherwise
dp[i] = books[i] # dp[i] is just the number of books at i
if i > 0: # if not the first book
dp[i] = max(dp[i], dp[i-1]) # compare with the previous book
return dp[n-1] # return the final answer

来自 newbing 的答案

This is a dynamic programming problem, where we can keep track of the maximum number of books we can take up to a given shelf i.

Let's define dp[i] as the maximum number of books we can take from shelves 0 to i (inclusive) while satisfying the condition that we must take strictly fewer books from shelf j than shelf j+1 for all j between 0 and i-1.

For each i, we need to find the maximum value of dp[j] + sum(books[j+1:i+1]) for 0 <= j < i, where sum(books[j+1:i+1]) represents the total number of books in the section from j+1 to i.

The reason why we use this formula is because we need to make sure that we take fewer books from shelf j compared to shelf j+1, so we need to add the sum of books from j+1 to i to dp[j]. We take the maximum value of this expression for all j between 0 and i-1 to get dp[i].

Finally, the answer will be the maximum value of dp[i] for all i between 0 and n-1.

Here's the Python code:

def max_books(books):
n = len(books)
dp = [books[0]] + [0]*(n-1)

for i in range(1, n):
for j in range(i):
if books[j] < books[i]:
dp[i] = max(dp[i], dp[j] + sum(books[j+1:i+1]))

return max(dp)

# Example usage:
print(max_books([8,5,2,7,9])) # Output: 19
print(max_books([7,0,3,4,5])) # Output: 12
print(max_books([8,2,3,7,3,4,0,1,4,3])) # Output: 13

上一条是来自开放猫的答案，懂得麻烦对比下，正确性如何

不对，大家可以用三个测试用例自己跑下测试下。chatGPT 给出的结果和跑出来的不等。
这个题要用 DP+单调栈去解，确实挺难的。

重新问 newbing ，并要求用 DP+单调栈去解，得到如下：

你给出的题目是 https://www.bing.com/search?form=SKPBOT&q=2355.%20Maximum%20Number%20of%20Books%20You%20Can%20Take ，这是一个动态规划和单调栈的结合问题。动态规划的思路是，对于每个位置 i ，我们维护一个 dp[i] 表示以 i 为右端点的区间能取到的最大书籍数。我们可以用一个单调栈来维护一个递减的序列，当我们遇到一个比栈顶元素小的数时，我们就找到了一个合法的区间，然后更新 dp[i] 为 dp[j] + (i - j) * books[i] ，其中 j 是栈顶元素的下标。这样我们就可以在 O(n) 的时间内求出所有的 dp[i] ，然后取最大值即可。我用 C++ 写了一段代码，你可以参考一下：

#include <iostream>
#include <vector>
#include <stack>
using namespace std;

int maxBooks(vector<int>& books, int l, int r) {
int n = books.size();
vector<int> dp(n + 1, 0); // dp[i] means the max books we can take from [0, i]
stack<int> st; // a monotonic decreasing stack
st.push(0); // a dummy element
int ans = 0;
for (int i = 1; i <= n; i++) {
while (!st.empty() && books[st.top()] > books[i]) {
// we find a valid interval [j + 1, i]
int j = st.top();
st.pop();
dp[i] = max(dp[i], dp[j] + (i - j) * books[i]);
}
st.push(i);
if (i >= l && i <= r) {
// update the answer if i is in the range
ans = max(ans, dp[i]);
}
}
return ans;
}

int main() {
vector<int> books = {8, 5, 2, 7, 9};
int l = 1, r = 4;
cout << maxBooks(books, l, r) << endl; // output: 19
return 0;
}

@jmk92 这个看起来靠谱，思路是对的。