1046. Shortest Distance (20)

题目链接：https://www.patest.cn/contests/pat-a-practise/1046
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>#include<algorithm>using namespace std;const int maxn=100010;int dis[maxn],A[maxn];int main(){    int N,M;    scanf("%d",&N);    int sum=0;    for(int i=1;i<=N;i++){        scanf("%d",&A[i]);        sum+=A[i];        dis[i]=sum;    }    scanf("%d",&M);    for(int i=0;i<M;i++){        int s,d;        scanf("%d %d",&s,&d);        if(d<s) swap(d,s);        int a=dis[d]-A[d]-(dis[s]-A[s]);        printf("%d\n",min(a,dis[N]-a));    }    return 0;}