1046 Shortest Distance （PAT甲级）

#include 
#include 
#include 

int N, M, d, u, v, sum, tmp;
std::vector dist;

int main(){
    scanf("%d", &N);
    sum = 0;
    dist.push_back(0);
    for(int i = 0; i < N; ++i){
        scanf("%d", &d);
        sum += d;
        dist.push_back(sum);
    }
    scanf("%d", &M);
    for(int i = 0; i < M; ++i){
        scanf("%d %d", &u, &v);
        if(u > v){
            std::swap(u, v);
        }
        tmp = dist[v - 1] - dist[u - 1];
        if(tmp * 2 > sum){
            tmp = sum - tmp;
        }
        printf("%d\n", tmp);
    }
    return 0;
}

题目如下：

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