# PAT A1046. Shortest Distance (20)

##### 题意重述

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.

Input Specification:

Each input file contains one test case. For each case, the first line contains an integer N (in [3, 10^5]), 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 (<=10^4), 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

##### 代码示例:
#include<cstdio>
using namespace std;
int main()
{
int N=0;
while(scanf("%d",&N)!=EOF)
{
long long D[100010]={0};
long long sum=0;
for(int i=1;i<=N;i++)
{
long long tmp1=0;
scanf("%lld",&(tmp1));
sum=sum+tmp1;
D[i]=sum;

}
int M=0;
scanf("%d",&M);
int left=0;
int right=0;
while(M--)
{
scanf("%d%d",&left,&right);
if(left>right)
{
int tmp=0;
tmp=left;
left=right;
right=tmp;
}
long long distance=0;
distance=D[right-1]-D[left-1];
if(distance<sum-distance)
{
printf("%lld\n",distance);
}
else
{
printf("%lld\n",sum-distance);
}
}
}
return 0;
}