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 ... 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 (<=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

思路

1.直接用累加求和，结果最后一项测试用例超时。
2.开一个二维数组暴力枚举出所有可能的情况，检测时直接输出对应情况，结果内存超了(最坏情况数组大小100001 * 100001）。
3.保存每一个目标点到起始点的road[i]，那么任意两个点s,e的距离可以通过road[s]-road[e]计算出来，而不用再去一步步累加，所以不会超时。而且这种解决方案最坏情况开辟的数组大小为100001，不会超内存。

代码

#include<iostream>

#include<vector>

#include<math.h>

using namespace std;

int main()

{

   int N;

   while(cin >> N)

   {

       vector<int> road(N + ,);

       int sum = ;

       for(int i = ;i <= N;i++)

       {

           int value;

           cin >> value;

           road[i + ] = road[i] + value;

           sum += value;

       }

        int M;

        cin >> M;

        while(M--)

        {

            int s,e;

            cin >> s >> e;

            int path = abs(road[s] - road[e]);

            cout << min(path,sum - path) << endl;

        }

   }

}