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

 #define _CRT_SECURE_NO_WARNINGS
 #include <climits>
 #include<iostream>
 #include<vector>
 #include<queue>
 #include<map>
 #include<set>
 #include<stack>
 #include<algorithm>
 #include<string>
 #include<cmath>
 using namespace std;
 int Dist[];
 int N, M;
 int Sum;
 void Ans(int i, int j)
 {
     int sum = ;
     if (i > j)Ans(j, i);
     else
     {
         for (int k = i; k < j; k++)
             sum += Dist[k];
         sum = (sum < (Sum - sum)) ? sum : Sum - sum;
         cout << sum << endl;
     }
 }
 int main()
 {

     cin >> N;
     for (int i = ; i <= N; i++)
     {
         cin >> Dist[i];
         Sum += Dist[i];
     }
     cin >> M;
     int v1, v2;
     for (int i = ; i < M; i++)
     {
         cin >> v1 >> v2;
         Ans(v1, v2);
     }
 }

ac代码

 #define _CRT_SECURE_NO_WARNINGS
 #include <climits>
 #include<iostream>
 #include<vector>
 #include<queue>
 #include<map>
 #include<set>
 #include<stack>
 #include<algorithm>
 #include<string>
 #include<cmath>
 using namespace std;
 vector<int>Dist;
 int sum;
 int main()
 {
     int N;
     cin >> N;
     Dist.resize(N + );
     for (int i = ; i <=N; i++)
     {
         int temp;
         cin >> temp;
         sum += temp;
         Dist[i] = sum;
     }
     int M;
     cin >> M;
     int v1, v2;
     for (int i = ; i < M; i++)
     {
         cin >> v1 >> v2;
         if (v1 > v2)
             swap(v1,v2);
         int s1 = Dist[v2 - ] - Dist[v1 - ];
         int s2 = sum - s1;
         if (s1 > s2)
             cout << s2 << endl;
         else
             cout << s1 << endl;
     }
 }