1089. Insert or Merge (25)【排序】——PAT (Advanced Level) Practise
2016-05-14 20:18
344 查看
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; }
相关文章推荐
- libdvbpsi源码分析(四)PAT表解析/重建
- PAT配置
- 什么是端口复用动态地址转换(PAT) 介绍配置实例
- MikroTik layer7-protocol
- PAT是如何工作的
- PAT 乙级题:1002. 写出这个数 (20)
- PAT (Advanced Level) Practise 1001-1010
- 数据结构学习与实验指导(一)
- PAT Basic Level 1001-1010解题报告
- PAT 数素数
- PAT 福尔摩斯的约会
- PAT 德才论
- PAT 月饼
- 1001. 害死人不偿命的(3n+1)猜想
- 1002. 写出这个数
- 1032. 挖掘机技术哪家强
- 1001. 害死人不偿命的(3n+1)猜想 (PAT basic)
- 1002. 写出这个数(PAT Basic)
- 1004. 成绩排名(PAT Basic)
- 1006. 换个格式输出整数(PAT Basic)