PAT甲级——A1046 Shortest Distance

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

 #include <iostream>
 #include <vector>
 using namespace std;
 int N, M;
 int main()
 {
     cin >> N;
     int num, a, b;
     vector<int>sum(N + , );
     for (int i = ; i <= N; ++i)
     {
         cin >> num;
         if (i == N)
             sum[] = sum[N] + num;
         else
             sum[i + ] = sum[i] + num;
     }
     cin >> M;
     for (int i = ; i < M; ++i)
     {
         cin >> a >> b;
         if (a > b)
             swap(a, b);
         int d1 = sum[b] - sum[a];
         int d2 = sum[] - sum[b] + sum[a] - sum[];
         cout << (d1 < d2 ? d1 : d2) << endl;
     }
     return ;
 }