Data hosted with ♥ by Pastebin.com - Download Raw - See Original
  1. Me:
  2. 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:
  3.  
  4. Problem Description: as it being stated, the description of the problem itself.
  5. Input: stating the constraint of the data set that will be passed to your solution.
  6. Output: stating the requirements that the results from your solution must satisfy.
  7. 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.
  8. Sample Output: that's the expected output your solution should work out for the Sample Input.
  9.  
  10. Do you understand?
  11.  
  12. ChatGPT:
  13. Yes, I understand. Please provide me with the problem description and I will try my best to provide you with a solution in C++.
  14.  
  15. Me:
  16. Here's the problem:
  17. Problem Description
  18. 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：
  19. 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。
  20.  
  21. Input
  22. 第一行是整数T，代表测试数据的组数（T ≤ 10）。
  23. 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。
  24. 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。
  25.  
  26. Output
  27. 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。
  28.  
  29. Sample Input
  30. 3
  31. 5
  32. 1 2 3 4 5
  33. 7
  34. 3 2 1 4 1 2 3
  35. 10
  36. 3 1 4 1 5 9 2 6 5 3
  37.  
  38. Sample Output
  39. 5
  40. 8
  41. 18
  42.  
  43. ChatGPT:
  44. Here is a solution in C++:
  45.  
  46. ```
  47. #include <iostream>
  48. #include <algorithm>
  49. #include <cstring>
  50.  
  51. using namespace std;
  52.  
  53. const int MAXN = 100005;
  54.  
  55. int n;
  56. int h[MAXN], s[MAXN];
  57.  
  58. int main() {
  59. int t;
  60. cin >> t;
  61.  
  62. while (t--) {
  63. cin >> n;
  64. memset(s, 0, sizeof(s));
  65.  
  66. for (int i = 1; i <= n; i++) {
  67. cin >> h[i];
  68. }
  69.  
  70. for (int i = 1; i <= n; i++) {
  71. s[i] = s[i-1] + h[i];
  72. }
  73.  
  74. int ans = 0, pre = 0;
  75. for (int i = 1; i <= n; i++) {
  76. ans += (s[i] - s[pre] >= i - pre);
  77. if (s[i] - s[pre] < i - pre) pre = i - 1;
  78. }
  79.  
  80. cout << ans << endl;
  81. }
  82.  
  83. return 0;
  84. }
  85. ```
  86.  
  87. Explanation:
  88.  
  89. 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.
  90.  
  91. 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`.
  92.  
  93. Finally, the answer is the number of sections that have been sprayed with material.
  94.  
  95. The time complexity of this solution is O(n), where n is the number of sections in the object.
  96.