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, 105]), 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 (<=104), 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 107.

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

晚上被這道題坑了很久，一直沒想明白為什麼不能完全ac,只能拿17/20的分，，我一開始的想法是每次都一個迴圈遍歷相加來求距離，查了網上資料說，在極端情況下，每次查詢都需要遍歷整個陣列，即有1e5次操作，而共有1e4次查詢，所以極端情況下會有1e9次操作，這在100ms內往往會超時。
解決辦法是一開始設定一個dis[i]

#include<cstdio>

using namespace std;

#define MAX 100001

int dis[MAX] , a[MAX];

int main(void){
    int sum = 0;
    int n,m,v1,v2;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++){
        scanf("%d",&a[i]);
        sum += a[i];
        dis[i] = sum;
    }
    
    scanf("%d",&m);
    while(m--){
        scanf("%d %d",&v1,&v2);
        if(v1 > v2){
            int t = v1;
            v1 = v2;
            v2 =t;
        }
        
        int ans = dis[v2-1] - dis[v1-1];
        if(ans > (sum-ans) )
            ans = sum -ans;
        printf("%d\n",ans);
    }

    
    return 0;
}