1046 Shortest Distance （20 分）
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 ⋯ 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 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
一，注意点
1，关于时间超时的问题，采取的措施是在输入的时候进行一个dis[]数组的计算。然后真要计算的时候就用dis[right-1]-dis[left-1]，而不是从头开始循环。
二，代码
#include<cstdio>
#include<algorithm>
using namespace std;
int arr[100010];
int dis[100010];
int main() {
int num = 0;
int sum = 0;
scanf("%d", &num);
for (int i = 1; i <= num; i++) {
scanf("%d", &arr[i]);
sum += arr[i];
dis[i]=sum;
}
int countting = 0;
int num_1=0,num_2=0;
int sum_1=0;
int temp=0;
scanf("%d", &countting);
for(int i=0;i<countting;i++){
sum_1=0;
scanf("%d %d",&num_1,&num_2);
if(num_1>num_2){
temp=num_1;
num_1=num_2;
num_2=temp;
}
sum_1=dis[num_2-1]-dis[num_1-1];
printf("%d\n",min(sum_1,sum-sum_1));
}
return 0;
}