[PAT] 1046 Shortest Distance （20 分）Java

The task is really simple: given N exits on a highway which forms a simple cycle, you are supposed to tell the shortest distance between any pair of exits.

Input Specification:

Each input file contains one test case. For each case, the first line contains an integer N (in [3,10​5​​]), followed by N integer distances D​1​​ D​2​​ ⋯ D​N​​, where D​i​​ is the distance between the i-th and the (i+1)-st exits, and D​N​​ is between the N-th and the 1st exits. All the numbers in a line are separated by a space. The second line gives a positive integer M (≤10​4​​), with M lines follow, each contains a pair of exit numbers, provided that the exits are numbered from 1 to N. It is guaranteed that the total round trip distance is no more than 10​7​​.

Output Specification:

For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

Sample Input:

5 1 2 4 14 9
3
1 3
2 5
4 1

Sample Output:

3
10
7

 1 package pattest;
 2 
 3 import java.io.BufferedReader;
 4 import java.io.IOException;
 5 import java.io.InputStreamReader;
 6 
 7 /**
 8  * @Auther: Xingzheng Wang
 9  * @Date: 2019/2/23 11:17
10  * @Description: pattest
11  * @Version: 1.0
12  */
13 public class PAT1046 {
14     public static void main(String[] args) throws IOException {
15         BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
16         String[] first = reader.readLine().split(" ");
17 
18         int n = Integer.parseInt(first[0]);
19         int[] dis = new int[n + 1];
20         int[] arr = new int[n + 1];
21         int sum = 0;
22 
23         for (int i = 1; i <= n; i++) {
24             dis[i] = Integer.parseInt(first[i]);
25             sum += dis[i];
26             arr[i] = sum;
27         }
28         int m = Integer.parseInt(reader.readLine());
29         for (int i = 0; i < m; i++) {
30             String[] line = reader.readLine().split(" ");
31             int s = Integer.parseInt(line[0]);
32             int e = Integer.parseInt(line[1]);
33             if (s > e) {
34                 int temp = s;
35                 s = e;
36                 e = temp;
37             }
38             int pre1 = arr[e - 1] - arr[s - 1];
39             int pre2 = sum - pre1;
40             System.out.println(Math.min(pre1,pre2));
41         }
42     }
43 }

标签：Distance,Java,1046,exits,10,int,each,new,distance
来源： https://www.cnblogs.com/PureJava/p/10498084.html