PAT甲级1046. Shortest Distance (20)

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

#include <cstdio>using namespace std;#include <vector>int main(){    int N;    scanf("%d",&N);    vector<int> V(N+1);    V[0]=0;    for(int i=1;i<=N;i++) {        scanf("%d",&V[i]);        V[i]+=V[i-1]; //V[i]表示 exit_1到exit_(i+1)的距离     }    int M;    scanf("%d",&M);    for(int i=0;i<M;i++){        int e1,e2;        scanf("%d %d",&e1,&e2);        if(e1>e2){            int tmp=e1;            e1=e2;            e2=tmp;        }        int d1=0,d2=0,d_min;        d1=V[e2-1]-V[e1-1];        d2=V[N]-d1;        d_min=d1<d2?d1:d2;        printf("%d\n",d_min);    }    return 0;}