1089. Insert or Merge (25)【排序】——PAT (Advanced Level) Practise
来源:互联网 发布:淘宝喵喵通信 编辑:程序博客网 时间:2024/04/20 19:54
题目信息
1089. Insert or Merge (25)
时间限制200 ms
内存限制65536 kB
代码长度限制16000 B
Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, and inserts it there. It repeats until no input elements remain.
Merge sort works as follows: Divide the unsorted list into N sublists, each containing 1 element (a list of 1 element is considered sorted). Then repeatedly merge two adjacent sublists to produce new sorted sublists until there is only 1 sublist remaining.
Now given the initial sequence of integers, together with a sequence which is a result of several iterations of some sorting method, can you tell which sorting method we are using?
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=100). Then in the next line, N integers are given as the initial sequence. The last line contains the partially sorted sequence of the N numbers. It is assumed that the target sequence is always ascending. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in the first line either “Insertion Sort” or “Merge Sort” to indicate the method used to obtain the partial result. Then run this method for one more iteration and output in the second line the resulting sequence. It is guaranteed that the answer is unique for each test case. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.
Sample Input 1:
10
3 1 2 8 7 5 9 4 6 0
1 2 3 7 8 5 9 4 6 0
Sample Output 1:
Insertion Sort
1 2 3 5 7 8 9 4 6 0
Sample Input 2:
10
3 1 2 8 7 5 9 4 0 6
1 3 2 8 5 7 4 9 0 6
Sample Output 2:
Merge Sort
1 2 3 8 4 5 7 9 0 6
解题思路
单步验证排序方式即可
AC代码
#include <cstdio>#include <algorithm>#include <vector>using namespace std;vector<int> a, b, c;void merge_sort(vector<int>& a, int b, int e, int step){ for (int i = b; i < e; i += step + step){ inplace_merge(a.begin() + i, a.begin() + min(e, i + step), a.begin() + min(e, i + step + step)); }}void insert_sort(vector<int>& a, int b, int e){ while (b + 1 < e && a[b] <= a[b + 1]) ++b; if (++b < e){ while (b > 0 && a[b] < a[b - 1]) { swap(a[b], a[b - 1]); --b; } }}int main(){ int n; scanf("%d", &n); c.resize(n); b.resize(n); for (int i = 0; i < n; ++i){ scanf("%d", &c[i]); } for (int i = 0; i < n; ++i){ scanf("%d", &b[i]); } a = c; bool flag = false; for (int i = 0; i < n; ++i){ if (a == b){ printf("Insertion Sort\n"); insert_sort(a, 0, n); flag = true; break; } insert_sort(a, 0, n); } if (!flag){ a = c; printf("Merge Sort\n"); for (int step = 1; step <= n; step += step){ merge_sort(a, 0, n, step); if (a == b){ merge_sort(a, 0, n, step + step); break; } } } printf("%d", a[0]); for (int i = 1; i < n; ++i){ printf(" %d", a[i]); } printf("\n"); return 0;}
- 1089. Insert or Merge (25)【排序】——PAT (Advanced Level) Practise
- PAT (Advanced Level) Practise 1089 Insert or Merge (25)
- PAT (Advanced Level) Practise 1089 Insert or Merge (25)
- 【PAT】【Advanced Level】1089. Insert or Merge (25)
- 1098. Insertion or Heap Sort (25)【排序】——PAT (Advanced Level) Practise
- 1029. Median (25)【排序】——PAT (Advanced Level) Practise
- PAT (Advanced Level) 1089. Insert or Merge (25) 插入和归并
- Pat(Advanced Level)Practice--1089(Insert or Merge )
- 1025. PAT Ranking (25)【排序】——PAT (Advanced Level) Practise
- 1075. PAT Judge (25)【排序】——PAT (Advanced Level) Practise
- 1033. To Fill or Not to Fill (25)【贪心+模拟】——PAT (Advanced Level) Practise
- 1028. List Sorting (25)【排序】——PAT (Advanced Level) Practise
- 1039. Course List for Student (25)【排序】——PAT (Advanced Level) Practise
- 1047. Student List for Course (25)【排序】——PAT (Advanced Level) Practise
- 1048. Find Coins (25)【排序】——PAT (Advanced Level) Practise
- 1052. Linked List Sorting (25)【链表+排序】——PAT (Advanced Level) Practise
- 1055. The World's Richest (25)【排序】——PAT (Advanced Level) Practise
- 1062. Talent and Virtue (25)【排序】——PAT (Advanced Level) Practise
- TSM备份相关概述
- equals与 "==" 的区别
- java hdu2011多项式求和
- Modular Multiplicative Inverse(模乘逆元)
- 文件管理器和文件连接器
- 1089. Insert or Merge (25)【排序】——PAT (Advanced Level) Practise
- linux centos lamp开启GD库的支持
- "围观"设计模式(10)--创建型之原型模式(Prototype Pattern)
- 微信JS SDK PHP Demo
- 分布式环境Session处理方法
- 1090. Highest Price in Supply Chain (25)【树】——PAT (Advanced Level) Practise
- int与Integer的区别
- php &$
- bestcoder百度之星2016AK 1001&1002&1003&1004 题解