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 D₁D₂ ... 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 14 931 32 54 1

Sample Output:

3107

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1. #include <iostream>
2. #include <vector>
3. #include <stdio.h>
4. #pragma warning(disable:4996)
6. using namespace std;
8. vector<int> num;
10. int main(void) {
11. int n,sum = 0;
12. cin >> n;
13. for (int i = 0; i < n; i++) {
14. int roadtemp;
15. scanf("%d", &roadtemp);
17. sum += roadtemp;
18. num.push_back(sum);
19. }
20. int m;
21. cin >> m;
23. for (int i = 0; i < m; i++) {
24. int start, end, change;
25. // cin >> start >> end;
26. scanf("%d %d", &start, &end);
27. if (start > end) {
28. change = end;
29. end = start;
30. start = change;
31. }
32. int road = 0;
34. if (start == 1) {
35. road = num[end-2];
36. }
37. else {
38. road = num[end-2] - num[start-2];
39. }
41. if (road > num[n-1] - road)
42. road = num[n-1] - road;
44. //cout << road << endl;
45. printf("%d\n", road);
46. }
48. return 0;
50. }