1046 Shortest Distance

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 D1 D2 ⋯ DN, where Di is the distance between the i-th and the (i+1)-st exits, and DN 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

#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;

int dis[100010] = { 0 };  //点1到点i之间顺时针的距离
int A[100010] = { 0 };  //点i到点i+1之间的距离

int main()
{
  int n = 0;  //点的个数
  int N = 0;  //要求的组数
  int sum = 0;
  int left = 0;
  int right = 0;

  scanf_s("%d", &n);
  for (int i = 1; i <= n; i++)
  {
    scanf_s("%d", &A[i]);
    sum += A[i];
    dis[i+1] = sum;
  }

  scanf_s("%d", &N);
  for (int i = 0; i < N; i++)
  {
    scanf_s("%d%d", &left, &right);
    if (left > right)
      swap(left, right);
    int min = (dis[right] - dis[left]) < sum - (dis[right] - dis[left]) ? dis[right] - dis[left] : sum - (dis[right] - dis[left]);
    printf("%d\n", min);
  }

}