题目如下：
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 D1D2… DN, where Diis the distance between the i-th and the (i+1)-st exits, and DNis 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 14 931 32 54 1Sample Output:3107
题目要求找出简单环路（类似表盘）中指定两点的最短距离，实质就是找出顺时针和逆时针两种路径中较小的，由于题目给出的数组规模比较庞大，因此不能简单的计算顺时针和逆时针的距离之和然后比较输出，而应当仔细思考问题的规律，试图不经过环的遍历就找到答案。
首先，我们可以在输入时累计相邻点的距离，得到环的总长sum，，从而只需要计算一个方向的距离，另一个方向一减便知。
其次，我们可以再用一个数组或vector记录从起点1顺时针走到当前结点的距离，在输入时就可以得到。
最后，有了这两个量，我们可以用较大结end点到1的距离减去较小结点start到1的距离，从而得到start到end的顺时针距离，逆时针距离就等于sum-顺时针距离，从而在不遍历环的情况下得到两个距离，这样才能避免最后一个测试点超时。
通过这个题目，我反思到对于简单的题目并不能掉以轻心，应当力图寻求事物的规律来替代繁琐的过程。
#include <iostream>#include <vector>#include <stdio.h>#include <math.h>using namespace std;vector<int> dis;vector<int> disToOrigin;int sum = 0;int computeMinDis(int s, int e){int minDist1 = 0,minDist2 = 0;int start, end;if(s < e) { start = s; end = e; }else if( s == e ) return 0;else { start = e; end = s; }minDist1 = disToOrigin[end] – disToOrigin[start];minDist2 = sum – minDist1;return minDist1 < minDist2 ? minDist1 : minDist2;}int main(){int N,M;cin >> N;dis.resize(N+1);disToOrigin.resize(N+1);disToOrigin[1] = 0;int value;for(int i = 1; i <= N; i++){scanf("%d",&value);dis[i] = value;if(i > 1) disToOrigin[i] = sum;sum += value;}cin >> M;int x,y;for(int i = 0; i < M; i++){scanf("%d%d",&x,&y);printf("%d\n",computeMinDis(x,y));}return 0;}
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