pat甲级—Shortest Distance (20)

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题目链接：https://www.nowcoder.com/pat/5/problem/4085

题目描述

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.

输入描述:

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.

输出描述:

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

输入例子:

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

输出例子:

3
10
7

n个数字围成一个环，求点A到点B的最短距离，可以有两个不同方向，取这两个方向最小的。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int main(){ 
	int n,m,st,en;
	int a[100005];
	int dis[100005];
	while(~scanf("%d",&n)){
		int sum=0;
		memset(dis,0,sizeof(dis));
		for(int i=1;i<=n;i++){
			scanf("%d",&a[i]);
			sum+=a[i];
			dis[i+1]=dis[i]+a[i];
		}
		scanf("%d",&m);
		for(int i=1;i<=m;i++){
			scanf("%d%d",&st,&en);
			if(st>en){
				swap(st,en);
			}
			int t=dis[en]-dis[st]; 
			printf("%d\n",min(t,sum-t));	
		}
	}
	return 0;
}