codeup 3.1简单模拟——最短距离

Description

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

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.

Output

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

题目大意：有一个环状高速公路，上面有给定的N个站点（出口），对于给定的M对站点，你要求出每对站点的最短距离（从顺时针和逆时针选最短路径），第一行表示站点之间的距离。

#include<iostream>
#include<cmath>
using namespace std;

int main(void)
{
    int n;
    while(cin>>n)
    {
        int distance[n],sum=0;
        for(int i=0;i<n;i++)
        {
            cin>>distance[i];
            sum+=distance[i];
        }
        int m,exit1,exit2;
        cin>>m;
        while(m--)
        {
            cin>>exit1>>exit2;
            int k=abs(exit2-exit1);
            int start=exit1;
            if(exit1>exit2) start=exit2;
            int result=0;
            for(int j=start-1;(k--)>0;j++)
            {
                result+=distance[j];
            }
            result=(result<=(sum-result)?result:(sum-result));
            std::cout << result << std::endl;
        }
    }
}