浙江大学PAT_甲级_1046. Shortest Distance (20)

题目链接：点击打开链接

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

我的C++程序：

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
int main()
{
	int n,m;//n个点,m对两点距离
	int temp,min_distance=0,cycle_length=0;//min_distance两点最短距离，cycle_length圈长度
	int a, b;//起点a,终点b
	int i = 0;
	int distance[100000];//istance[i]存储第i个点到第一个点的距离，i>=1
	cin >> n;
	for (i = 1;i <= n;i++)
	{
		cin >> temp;
		cycle_length += temp;//计算圆圈长度
		distance[i + 1] = distance[i] + temp;
	}
	cin >> m;
	while (m--)
	{
		cin >> a >> b;
		min_distance = abs(distance[b] - distance[a]);
		cout << min(min_distance, cycle_length- min_distance)<< endl;//圆圈内的两点最短距离
	}
	//system("pause");
	return 0;
}