A1046. Shortest Distance (20)

 

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 9
3
1 3
2 5
4 1

Sample Output:

3
10
7

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
using namespace std;



const int maxn=100005; 
int main(){
    int dis[maxn]={0},a[maxn]={0};
    int sum=0,query,n,left,right;
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);//从0开始计数  
        sum+=a[i];//a[i]表示i-1到i之间的距离 
        dis[i]=sum;//dis[i]表示0->i之间的距离。 
    }
    scanf("%d",&query);
    for(int i=0;i<query;i++)
    {
        scanf("%d %d",&left,&right);
        if(left>right)swap(left,right);
        int temp=dis[right-1]-dis[left-1];
        printf("%d
",min(temp,sum-temp));
    } 
 
   
    return 0;
}