题目地址

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 <vector>
#include<algorithm>
#include <cmath>
#include<map>
#include<cstring>
#include<queue>
#include<string>
#include<set>
using namespace std;
typedef long long ll;
const int maxn=100010,inf=100000000;
int main(){
    int sum=0,dis[maxn],n,k,u,v;
    cin>>n;
    for(int i=1;i<=n;i++) {
        cin>>dis[i];sum+=dis[i];
        if(i!=1) dis[i]=dis[i]+dis[i-1];
    }
    cin>>k;
    while(k--){
        cin>>u>>v;
        if(u>v) swap(u,v);
        cout<<min(dis[v-1]-dis[u-1],sum-(dis[v-1]-dis[u-1]))<<endl;
    }
}