1046 Shortest Distance (20 分)

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 [3,10​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 #include <iostream> #include<stdio.h> #include<math.h> using namespace std; const int maxn=100001; int dis[maxn]; int main() { int n; int sum=0; scanf("%d",&n); for(int i=1; i<=n; i++) { int t; scanf("%d",&t); sum+=t; dis[i]=sum; } int m; scanf("%d",&m); while(m--) { int x,y; scanf("%d%d",&x,&y); if(x>y) swap(x,y); // printf("%d %d\n",dis[y],dis[x]); int ans=dis[y-1]-dis[x-1]; printf("%d\n",min(sum-ans,ans)); } return 0; }