PAT甲级无算法1046 Shortest Distance

现在一个很大的问题就是容易把问题复杂化，而正解往往又是在原来基础上加以简化得出的。一看就懂，确实最难的，这两天的做题都是如此。注意不要闭门造车。

 

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 [3,10​5​​]), followed by N integer distances D​1​​ D​2​​ ⋯ 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 (≤10​4​​), 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 10​7​​.

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

看到一个很好的代码，原样贴下。原帖移步https://www.cnblogs.com/zlrrrr/p/9439663.html

#include <bits/stdc++.h>
using namespace std;
 
const int maxn = 1e5 + 10;
int a[maxn], dis[maxn];
 
int main() {
    int N, sum = 0;
    scanf("%d", &N);
    for(int i = 1; i <= N; i ++) {
        scanf("%d", &a[i]);
        sum += a[i];   //在输入过程中建立dis数组
        dis[i] = sum;  //降低时间复杂度
    }
 
    int T, left, right, temp;
    scanf("%d", &T);
    for(int i = 1; i <= T; i ++) {
        scanf("%d%d", &left, &right);
        if(left > right) {
            swap(left, right);   //swap函数交换
        }
        temp = dis[right - 1] - dis[left - 1];
        printf("%d\n", min(temp, sum - temp));//min函数判断
    }
 
    return 0;
 }