1046 Shortest Distance (20分)

高速公路出口距离计算

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本文介绍了一个简单的算法，用于计算高速公路形成简单环形时任意两个出口之间的最短距离。输入包括出口数量、各出口间的距离及若干对出口编号，输出为每对出口间的最短距离。

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

#include<iostream>
#include<vector>

using namespace std;

int main(){
    int n;
    cin >> n;
    vector<int> vec(n + 1);
    int sum = 0;
    for(int i = 1; i <= n; i++){
        scanf("%d", &vec[i]);
        sum += vec[i];
    }
    int m;
    cin >> m;
    for(int i = 0; i < m; i++){
        int a, b;
        scanf("%d %d", &a, &b);
        int temp = 0;
        if(a > b){

            for(int j = b; j < a; j++){
                temp += vec[j];
            }   

        }else
        {
            for(int j = a; j < b; j++){
                temp += vec[j];
            } 
            
        }
        
        
        if(temp > sum - temp){
            printf("%d\n", sum - temp);        
        }else
        {
            printf("%d\n", temp);  
        }
        

    }

    return 0;
}