1046 Shortest Distance (20 分)
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 ⋯ DN, where Di is the distance between the i-th and the (i+1)-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 (≤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
#include<bits/stdc++.h>
using namespace std;
const int Max = 100010;
int main(){
int n, x, dis[Max];
scanf("%d", &n);
fill(dis, dis + Max, 0);
for(int i = 1; i <= n; i++){
scanf("%d", &x);
dis[i] = dis[i - 1] + x;
}
int m;
scanf("%d", &m);
int start, end;
for(int i = 0; i < m; i++){
scanf("%d %d", &start, &end);
if(start > end)
swap(start, end);
int a = dis[end - 1] - dis[start - 1];
int b = dis[n] - a;
x = min(a, b);
printf("%d\n", x);
}
}