1046 Shortest Distance （20 分）
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 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
//本題感覺有點崩，做了半下午……瘋狂拉低通過率
//一個環，給環上兩點算最短路徑長度，即順時針和逆時針的路徑長度最小值
//我們只需計算其中一個方向上的，然後用環的總長度減去它，即得逆向的長度
//時間複雜度有要求，總的時間複雜度是O(N)吧，之前寫了兩三個算法，都是最後一個測試點超時
#include <cstdio>
#include <cstring>
#include <cmath>
int D[100001] = { 0 };//各地點間的距離
int main()
{
int N = 0, M = 0;
scanf("%d", &N);
for (int i = 1, dis = 0; i <= N; i++)
{
scanf("%d", &dis);
D[i] += D[i - 1] + dis;//累加路程長度
}
scanf("%d", &M);
for (int i = 0; i < M; i++)
{
int a = 0, b = 0, path1 = 0, path2 = 0;
scanf("%d %d", &a, &b);
if (a > b) path1 = D[a-1] - D[b-1];//先計算出正向的長度；這裏很巧妙
else path1 = D[b-1] - D[a-1];
path2 = D[N] - path1;//反向長度即總長度減去正向的長度（環形）
printf("%d\n", path1 < path2 ? path1 : path2);//輸出較小者即可
}
return 0;
}