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]), followed by Ninteger distances D1 D2 ⋯ DN, where Di is the distance between the i-th and the (-st exits, and DN 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(≤), 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 1.
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
水题，就是一个前缀和就完事。
1 #include <bits/stdc++.h> 2 using namespace std; 3 int n,m; 4 int an[100005]; 5 int main(){ 6 cin >> n; 7 int sum = 0; 8 for(int i = 1; i <= n ; i ++){ 9 cin >> an[i]; 10 sum += an[i]; 11 an[i] += an[i-1]; 12 } 13 cin >> m; 14 int xx, yy, x, y; 15 for(int i = 0; i < m; i ++){ 16 cin >> xx >> yy; 17 y = max(xx,yy); 18 x = min(xx,yy); 19 int cnt = an[y-1]-an[x-1]; 20 int cnt1 = sum - cnt; 21 int ans = min(cnt, cnt1); 22 cout << ans << endl; 23 } 24 return 0; 25 }