PAT甲级——A1046 Shortest Distance

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]), followed by N integer distances D1 D2 ⋯ DN, where Di is the distance between the i-th and the (-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 (≤), 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 1.

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:

 #include <iostream>

 #include <vector>

 using namespace std;

 int N, M;

 int main()

 {

     cin >> N;

     int num, a, b;

     vector<int>sum(N + , );

     for (int i = ; i <= N; ++i)

     {

         cin >> num;

         if (i == N)

             sum[] = sum[N] + num;

         else

             sum[i + ] = sum[i] + num;

     }

     cin >> M;

     for (int i = ; i < M; ++i)

     {

         cin >> a >> b;

         if (a > b)

             swap(a, b);

         int d1 = sum[b] - sum[a];

         int d2 = sum[] - sum[b] + sum[a] - sum[];

         cout << (d1 < d2 ? d1 : d2) << endl;

     }

     return ;

 }

巴特西

PAT甲级——A1046 Shortest Distance

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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