1046. Shortest Distance

1046. Shortest Distance (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

 1 #include<stdio.h>
 2 #include<math.h>
 3 #include<stdlib.h>
 4 #include<string.h>
 5 int dis[100010];
 6 int main()
 7 {
 8     int n, m, i, j, d, sum = 0;
 9     scanf("%d", &n);
10     for(i = 1; i <= n; i++)
11     {
12         scanf("%d", &d);
13         sum += d;
14         dis[i] = sum;
15     }
16     scanf("%d", &m);
17     for(i = 0; i < m; i++)
18     {
19         int x, y;
20         scanf("%d%d", &x, &y);
21         if(x > y)
22         {
23             int temp = y;
24             y = x;
25             x = temp;
26         }
27         int a1 = dis[y - 1] - dis[x - 1], a2 = sum - a1;
28         if(a1 > a2)
29         {
30             int temp2 = a1;
31             a1 = a2;
32             a2 = temp2;
33         }
34         printf("%d
", a1);
35     }
36     return 0;
37 }