PAT 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, 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 14 931 32 54 1

Sample Output:

310
7
题解
#include<cstdio>#include<iostream>#include<vector>#include<algorithm>using namespace std;int main() {vector<long>a, dis;long sum = 0, temp, m, n, left, right;cin >> n;dis.push_back(0);for (int i = 0; i < n; i++) {cin >> temp;sum += temp;;dis.push_back(sum);}cin >> m;for (int i = 0; i < m; i++) {cin >> left >> right;if (left > right)swap(left, right);temp = dis[right - 1] - dis[left - 1];temp = min(temp, sum - temp);a.push_back(temp);}for (int i = 0; i < m; i++) {cout << a[i] << endl;}return 0;}//此题要注意的是如果不处理dis和sum的话每次对数组进行遍历//极有可能超出运行时间，另外要注意的就是距离的对应关系