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<iostream> #include<cmath> #include<string.h> #include<algorithm> #include<cstdio> #include<queue> #include<set> #include<map> #include<vector> #include<string> using namespace std; typedef long long ll; 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++) { cin>>a[i]; sum+=a[i]; dis[i] = sum; } int T,left,right,temp; cin>>T; for(int i = 1; i <= T;i++) { cin>>left>>right; if(left>right){ swap(left,right); } temp = dis[right-1]-dis[left-1]; printf("%d\n",min(temp,sum-temp)); } }