1046 Shortest Distance (20point(s))

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 D1 D2⋯ DN, where D​i 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 (≤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

超时代码

#include<bits/stdc++.h>
using namespace std;
int d[100000+10];
int main(){
    int n,m;
    int start,end;
    cin>>n;
    for(int i=1;i<=n;++i){
        cin>>d[i];
    }
    cin>>m;
    for(int i=0;i<m;++i){
        cin>>start>>end;
        if(start>end)
            swap(start,end);
        int minDis=0,temp=0;
        for(int k=start;k<end;++k){
            temp+=d[k];
        }
        for(int k=start-1;k>=1;--k)
            minDis+=d[k];
        for(int k=end;k<=n;++k)
            minDis+=d[k];
        minDis=min(minDis,temp);
        cout<<minDis<<endl;
    }
}

打表

#include<bits/stdc++.h>
using namespace std;
int d[100000+10];
int main(){
    int n,m;
    int start,end;
    cin>>n;
    int temp,sum=0;
    for(int i=1;i<=n;++i){
        cin>>temp;
        sum+=temp;
        d[i]=sum;
    }
    cin>>m;
    for(int i=0;i<m;++i){
        cin>>start>>end;
        if(start>end)
            swap(start,end);
        int t=d[end-1]-d[start-1];
        cout<<min(t,sum-t)<<endl;
    }
}