A1046

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<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<cstdio>
#include<algorithm>
using namespace std;
int main(){
    int N,num[100005],dis[100005],M,a,b,sum=0;
    cin>>N;
    for(int i=1;i<=N;i++){
        scanf("%d",&num[i]);
        sum+=num[i];
        dis[i]=sum;
    }
    cin>>M;
    while(M--){
        scanf("%d%d",&a,&b);
        if(a>b)swap(a,b);
        int temp=dis[b-1]-dis[a-1];
        printf("%d\n",min(temp,sum-temp));
    }
    return 0;
}