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 [3,10​5​​]), followed by N integer distances D​1​​ D​2​​ ⋯ D​N​​, where D​i​​ is the distance between the i-th and the (i+1)-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 (≤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<iostream>
#include<cmath>
#include<string.h>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<set>
#include<map>
#include<vector>
#include<string>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
int a[maxn],dis[maxn];
int main(){
    int N,sum = 0;
    scanf("%d",&N);
    for(int i = 1; i <= N ;i++)
    {
        cin>>a[i];
        sum+=a[i];
        dis[i] = sum;
    }
    int T,left,right,temp;
    cin>>T;
    for(int i = 1; i <= T;i++)
    {
        cin>>left>>right;
        if(left>right){
            swap(left,right);
        }
        temp = dis[right-1]-dis[left-1];
        printf("%d\n",min(temp,sum-temp));

    }
}