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 }
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 }
原文地址：https://www.cnblogs.com/57one/p/12037128.html