题目描述

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

注意点

一般这种短小精悍的题都会卡时限，因此不要大意，一开始就要朝着缩短时间的思路去。cin cout的耗时远远大于scanf printf，因此如果有测试点通不过可以考虑改成scanf printf，在数据量较大的情况下，可以节省至少10ms。

代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
    int n, m, sum = 0, a, b, ans = 0, temp;
    scanf("%d", &n);
    vector<int> dis(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> temp;
        sum += temp;
        dis[i] = sum;
    }
    scanf("%d", &m);
    for (int i = 0; i < m; i++) {
        scanf("%d %d", &a, &b);
        if (a > b) swap(a, b);
        temp = dis[b - 1] - dis[a - 1];
        printf("%d\n", min(temp, sum - temp));
    }
    return 0;
}