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 原题链接：https://www.patest.cn/contests/pat-a-practise/1046
思路：
每一位存累加和，然后计算两个方向
CODE：
#include<iostream> using namespace std; int main() { int n; cin>>n; int sum[n+1]; sum[0]=0; for (int i=1;i<=n;i++) { cin>>sum[i]; sum[i]+=sum[i-1]; } int m; cin>>m; for (int i=0;i<m;i++) { int a,b; cin>>a>>b; if (a>b) swap(a,b); cout<<min(sum[b-1]-sum[a-1],sum[a-1]-sum[0]+sum[n]-sum[b-1])<<endl; } return 0; }