題目描述：
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.

輸入描述：

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.

輸出描述：

For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

輸入例子：

5 1 2 4 14 9
3
1 3
2 5
4 1

輸出例子：

3
10
7

思路：
　　題目給了我們N個點之間的距離，這N個點構成了一個環，求給出的幾組點之間的最短距離。
　　一開始我想到的是建立一個N*2的陣列，每行表示一個點到上個點和下個點的距離，但後來發現由於這是一個環，所以兩點之間有兩種走法，只要求出其中一種的距離，即可得另
　　一個。有了這個想法，寫出了很粗糙的第一版

 

 1 #include <iostream>
 2 using namespace std;
 3 
 4 int f(int a,int b,int array[],int sum){  //判斷兩點之間，哪條路徑短
 5     int m,n,left=0,right=0;  //right->從較小序號點向較大序號點的距離，left->另一條路徑距離
 6     if(a-1<b-1){  //防止a點的序號比b點大
 7         m=a-1;
 8         n=b-1;
 9     }else{
10         m=b-1;
11         n=a-1;
12     }
13     for(int i=m;i<n;++i)
14     {
15         right+=array[i];
16     }
17     left=sum-right;
18     return left<right?left:right;
19 }
20 
21 int main()
22 {
23     int N;
24     cin>>N;
25     int array_1[N],sum=0;  //array_1->存放每個點到下個點的距離，sum->儲存距離和
26     for(int i=0;i<N;++i)
27     {
28         cin>>array_1[i];
29         sum+=array_1[i];
30     }
31     int M;  //M->共M組點
32     cin>>M;
33     int array_2[M][2];
34     for(int i=0;i<M;++i)
35     {
36         cin>>array_2[i][0]>>array_2[i][1];
37     }
38     for(int i=0;i<M;++i)
39     {
40         cout<<f(array_2[i][0],array_2[i][1],array_1,sum)<<endl;
41     }
42 }