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 Ninteger 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

分析： 题目意思是给了一个环，已经环的每条边（公路）的长度，现给出起点和终点，求出起点和终点之间的最短路径长度，即求顺时针和逆时针行驶的距离最小值。

先求顺时针的距离，而逆时针距离=sum-顺时针距离；

另外暴力求解测试点3会超时，因为每次遍历数组，最大情况10^5个顶点，若有10^4个查询，则一共10^9，对于100ms的时限不能承受。

可以在输入距离时用dis数组记录从1顺时针出发到各顶点的距离。

 /**
 * Copyright(c)
 * All rights reserved.
 * Author : Mered1th
 * Date : 2019-02-23-19.12.21
 * Description : A1046
 */
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<unordered_set>
 #include<map>
 #include<vector>
 #include<set>
 using namespace std;
 ;
 int a[maxn],dis[maxn];
 int main(){
 #ifdef ONLINE_JUDGE
 #else
     freopen("1.txt", "r", stdin);
 #endif
     ;
     scanf("%d",&n);
     ;i<=n;i++){
         scanf("%d",&a[i]);
         sum+=a[i];
         dis[i]=sum;
     }
     scanf("%d",&m);
     ;i<m;i++){
         ,ans2=;
         scanf("%d%d",&st,&ed);
         if(st>ed) swap(st,ed);
         ans1=dis[ed-]-dis[st-];
         ans2=sum-ans1;
         printf("%d\n",min(ans1,ans2));
     }
     ;
 }