1046 Shortest Distance (20分)

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]), followed by N integer distances D​1​​ D​2​​ ⋯ D​N​​, where D​i​​ is the distance between the i-th and the (-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 (≤), 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 1.

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

题目分析：刚开始的直接暴力做 第三个点没过 看了别人的博客 认识了前缀数组
错误代码

 1 #define _CRT_SECURE_NO_WARNINGS
 2 #include <climits>
 3 #include<iostream>
 4 #include<vector>
 5 #include<queue>
 6 #include<map>
 7 #include<set>
 8 #include<stack>
 9 #include<algorithm>
10 #include<string>
11 #include<cmath>
12 using namespace std;
13 int Dist[10000];
14 int N, M;
15 int Sum;
16 void Ans(int i, int j)
17 {
18     int sum = 0;
19     if (i > j)Ans(j, i);
20     else
21     {
22         for (int k = i; k < j; k++)
23             sum += Dist[k];
24         sum = (sum < (Sum - sum)) ? sum : Sum - sum;
25         cout << sum << endl;
26     }
27 }
28 int main()
29 {
30 
31     cin >> N;
32     for (int i = 1; i <= N; i++)
33     {
34         cin >> Dist[i];
35         Sum += Dist[i];
36     }
37     cin >> M;
38     int v1, v2;
39     for (int i = 0; i < M; i++)
40     {
41         cin >> v1 >> v2;
42         Ans(v1, v2);
43     }
44 }

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ac代码

 1 #define _CRT_SECURE_NO_WARNINGS
 2 #include <climits>
 3 #include<iostream>
 4 #include<vector>
 5 #include<queue>
 6 #include<map>
 7 #include<set>
 8 #include<stack>
 9 #include<algorithm>
10 #include<string>
11 #include<cmath>
12 using namespace std;
13 vector<int>Dist;
14 int sum;
15 int main()
16 {
17     int N;
18     cin >> N;
19     Dist.resize(N + 1);
20     for (int i = 1; i <=N; i++)
21     {
22         int temp;
23         cin >> temp;
24         sum += temp;
25         Dist[i] = sum;
26     }
27     int M;
28     cin >> M;
29     int v1, v2;
30     for (int i = 0; i < M; i++)
31     {
32         cin >> v1 >> v2;
33         if (v1 > v2)
34             swap(v1,v2);
35         int s1 = Dist[v2 - 1] - Dist[v1 - 1];
36         int s2 = sum - s1;
37         if (s1 > s2)
38             cout << s2 << endl;
39         else
40             cout << s1 << endl;
41     }
42 }

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