A1046. Shortest Distance (20)

A1046

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

---

我的AC代码如下;

#include <cstdio>
const int maxn = 100005;
int dis[maxn], sum, n;
int main() {
    scanf("%d", &n);
    int num;
    dis[0] = 0;
    sum = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &num);
        dis[i] = dis[i-1]+num;
        sum += num;
    }
    int m, left, right;
    int ans, n_ans;
    scanf("%d", &m);
    while (m--) {
        scanf("%d%d", &left, &right);
        if (left > right) {
            int tmp = left;
            left = right;
            right = tmp;
        }
        ans = dis[right - 1] - dis[left - 1];
        n_ans = sum - ans;
        if (ans > n_ans) {
            ans = n_ans;
        }
        printf("%d\n", ans);
    }
    return 0;
}

注：一段一段保存总和，例如dis[1]是从第0个点到第1个点的距离，而dis[2]是从第0个点开始到第2个点的距离。

所以第1个点到第2个点的距离是dis[2]-dis[1];

但是由于输入时是从1开始计入的，所以是dis[left-1]-dis[right-1]