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]), followed by N integer distances D​1​​ D​2​​ ⋯ D​N​​, where D​i​​ is the distance between the i-th and the (-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 (≤), 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 1.

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

题目分析：刚开始的直接暴力做 第三个点没过 看了别人的博客 认识了前缀数组
错误代码

#define _CRT_SECURE_NO_WARNINGS
#include <climits>
#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
int Dist[10000];
int N, M;
int Sum;
void Ans(int i, int j)
{
    int sum = 0;
    if (i > j)Ans(j, i);
    else
    {
        for (int k = i; k < j; k++)
            sum += Dist[k];
        sum = (sum < (Sum - sum)) ? sum : Sum - sum;
        cout << sum << endl;
    }
}
int main()
{

    cin >> N;
    for (int i = 1; i <= N; i++)
    {
        cin >> Dist[i];
        Sum += Dist[i];
    }
    cin >> M;
    int v1, v2;
    for (int i = 0; i < M; i++)
    {
        cin >> v1 >> v2;
        Ans(v1, v2);
    }
}

ac代码

#define _CRT_SECURE_NO_WARNINGS
#include <climits>
#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
vector<int>Dist;
int sum;
int main()
{
    int N;
    cin >> N;
    Dist.resize(N + 1);
    for (int i = 1; i <=N; i++)
    {
        int temp;
        cin >> temp;
        sum += temp;
        Dist[i] = sum;
    }
    int M;
    cin >> M;
    int v1, v2;
    for (int i = 0; i < M; i++)
    {
        cin >> v1 >> v2;
        if (v1 > v2)
            swap(v1,v2);
        int s1 = Dist[v2 - 1] - Dist[v1 - 1];
        int s2 = sum - s1;
        if (s1 > s2)
            cout << s2 << endl;
        else
            cout << s1 << endl;
    }
}