PAT 1046. Shortest Distance (20)

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 ... D_N, where D_i is the distance between the i-th and the (i+1)-st exits, and D_N 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 <iostream>

using namespace std;

int exitDst[100000];
int sumDst[100001];

int main()
{
    int exitNum, totalDst=0;
    cin >> exitNum;
    for (int i = 0; i < exitNum; i++)
        scanf("%d", &exitDst[i]);
    for (int i = 1; i < exitNum; i++)
        sumDst[i] = sumDst[i - 1] + exitDst[i - 1];
    totalDst= sumDst[exitNum-1] + exitDst[exitNum-1];

    int queryNum;
    cin >> queryNum;
    for (int i = 0; i < queryNum; i++)
    {
        int start, end, Dst=0;
        scanf("%d%d", &start, &end);
        Dst = sumDst[end - 1] - sumDst[start - 1];
        Dst = Dst >0 ? Dst : -Dst;
        if (totalDst - Dst < Dst)
            printf("%d\n", totalDst - Dst);
        else
            printf("%d\n", Dst);
    }
}