題目

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

理解

計算一個環路上的兩個節點，順逆時針兩個方向哪邊路徑更短。

思路

在輸入時可以用sum變數更新總長度，這樣後邊計算可以只計算單側長度，然後用sum做減法，更加方便。千萬別忘了每次輸出要換行！！！

程式碼

#include<iostream>
using namespace std;
int main()
{
    int n=0;
    cin>>n;
    int sum=0;
    int *a=new int[n];
    for(int i=0; i<n; i++)
    {
        cin>>a[i];
        sum+
 
=a[i];
    }
    int k=0;
    cin>>k;
    for(int i=0; i<k; i++)
    {
        int p,q;
        cin>>p;
        cin>>q;
        p-=1;
        q-=1;
        if(p>q)
        {
            int temp=q;
            q=p;
            p=temp;
        }
        int r1=0,r2=0;
        for(int j=p; j<q; j++)
        {
            r1+=a[j];
        }
        r2=sum-r1;
        //不要忘記輸出換行！！！！！
        (r1<r2)?cout<<r1<<endl:cout<<r2<<endl;
    }
    return 0;
}