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- Me:
- I have a problem written in Chinese, you must understand it and work out a proper solution in C++. Note that the problem consisted of 5 sections, meaning of every parts are explained as following:
- Problem Description: as it being stated, the description of the problem itself.
- Input: stating the constraint of the data set that will be passed to your solution.
- Output: stating the requirements that the results from your solution must satisfy.
- Sample Input: you're not conscious of all the data set, so it's just a sample show you what kind of data will be passed to your solution.
- Sample Output: that's the expected output your solution should work out for the Sample Input.
- Do you understand?
- ChatGPT:
- Yes, I understand. Please provide me with the problem description and I will try my best to provide you with a solution in C++.
- Me:
- Here's the problem:
- Problem Description
- Zyj发明了一台3D打印机,但他的技术不过关,这台打印机的喷嘴频繁喷喷停停的话是会坏掉哒,所以Zyj为了尽量保证喷嘴能够尽量连续地喷出物料,设计了一个很蠢的方法:
- 一个打印物件可以大致分为n个部分,每个部分要求达到高度hi。每次可以选择从第L部分到第R部分(L≤R)喷上一层物料,可以增加高度1。请你帮助他算一算,对于给定的打印物件,至少需要喷涂多少次物料。
- Input
- 第一行是整数T,代表测试数据的组数(T ≤ 10)。
- 每组数据中,第一行是整数n(1≤n≤105),表示打印物件有几个部分。
- 接下来第二行包含n个用空格间隔的整数,每个整数hi(0≤hi≤105)代表该部分需求的高度。
- Output
- 对于每组数据,输出需要喷涂的最少次数,每个结果占据一行。
- Sample Input
- 3
- 5
- 1 2 3 4 5
- 7
- 3 2 1 4 1 2 3
- 10
- 3 1 4 1 5 9 2 6 5 3
- Sample Output
- 5
- 8
- 18
- ChatGPT:
- Here is a solution in C++:
- ```
- #include <iostream>
- #include <algorithm>
- #include <cstring>
- using namespace std;
- const int MAXN = 100005;
- int n;
- int h[MAXN], s[MAXN];
- int main() {
- int t;
- cin >> t;
- while (t--) {
- cin >> n;
- memset(s, 0, sizeof(s));
- for (int i = 1; i <= n; i++) {
- cin >> h[i];
- }
- for (int i = 1; i <= n; i++) {
- s[i] = s[i-1] + h[i];
- }
- int ans = 0, pre = 0;
- for (int i = 1; i <= n; i++) {
- ans += (s[i] - s[pre] >= i - pre);
- if (s[i] - s[pre] < i - pre) pre = i - 1;
- }
- cout << ans << endl;
- }
- return 0;
- }
- ```
- Explanation:
- The problem requires us to find the minimum number of times that the printer needs to spray material on a 3D object. We can approach this problem by using the prefix sum technique.
- Firstly, we calculate the prefix sum of the height of each section of the object, denoted by `s`. Then, we iterate through each section of the object from left to right, and maintain a variable `pre` that represents the last section that has been sprayed with material. If the difference between `s[i]` and `s[pre]` is greater than or equal to the distance between the current section and the last sprayed section, we can spray material on the current section, and update `pre` to `i-1`.
- Finally, the answer is the number of sections that have been sprayed with material.
- The time complexity of this solution is O(n), where n is the number of sections in the object.
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