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 D1 D2 ⋯ DN, where Di is the distance between the i-th and the (-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 (≤), 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[];

 int N, M;

 int Sum;

 void Ans(int i, int j)

 {

     int sum = ;

     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 = ; i <= N; i++)

     {

         cin >> Dist[i];

         Sum += Dist[i];

     }

     cin >> M;

     int v1, v2;

     for (int i = ; 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 + );

     for (int i = ; i <=N; i++)

     {

         int temp;

         cin >> temp;

         sum += temp;

         Dist[i] = sum;

     }

     int M;

     cin >> M;

     int v1, v2;

     for (int i = ; i < M; i++)

     {

         cin >> v1 >> v2;

         if (v1 > v2)

             swap(v1,v2);

         int s1 = Dist[v2 - ] - Dist[v1 - ];

         int s2 = sum - s1;

         if (s1 > s2)

             cout << s2 << endl;

         else

             cout << s1 << endl;

     }

 }

巴特西

1046 Shortest Distance (20分)

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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