The task is really simple: givenNexits 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 integerN(in [3]), followed byNinteger distancesD​1​​D​2​​⋯D​N​​, whereD​i​​is the distance between thei-th and the(-st exits, andD​N​​is between theN-th and the 1st exits. All the numbers in a line are separated by a space. The second line gives a positive integerM(≤), withMlines follow, each contains a pair of exit numbers, provided that the exits are numbered from 1 toN. It is guaranteed that the total round trip distance is no more than1.

Output Specification:

For each test case, print your results inMlines, 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

思路：用陣列每個點到第一個點的距離，把時間複雜度控制在O（1），注意：O（n）的時間複雜度會執行超時

#include<bits/stdc++.h>
using
 
 namespace std;
const int maxn=1000010;

int nums[maxn];
int dist[maxn];
int main(){
    int n,m;
    cin>>n;
    int sum=0;
    dist[0]=0;
    for(int i=1;i<=n;i++){
        cin>>nums[i];
        dist[i]=dist[i-1]+nums[i];
        sum+=nums[i];
    }
    cin>>m;
    int a,b;
    for(int
 
 i=0;i<m;i++){
        cin>>a>>b;
        int sum2=0;
        if(a>b){
            swap(a,b);
        }
        sum2=dist[b-1]-dist[a-1];
        cout<<min(sum-sum2,sum2)<<endl;
        
    }
    return 0;
}