标签：names ase really secure follow limit each res close
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.
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.
For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.
5 1 2 4 14 9
3
1 3
2 5
4 1
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
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <climits> 3 #include<iostream> 4 #include<vector> 5 #include<queue> 6 #include<map> 7 #include<set> 8 #include<stack> 9 #include<algorithm> 10 #include<string> 11 #include<cmath> 12 using namespace std; 13 int Dist[10000]; 14 int N, M; 15 int Sum; 16 void Ans(int i, int j) 17 { 18 int sum = 0; 19 if (i > j)Ans(j, i); 20 else 21 { 22 for (int k = i; k < j; k++) 23 sum += Dist[k]; 24 sum = (sum < (Sum - sum)) ? sum : Sum - sum; 25 cout << sum << endl; 26 } 27 } 28 int main() 29 { 30 31 cin >> N; 32 for (int i = 1; i <= N; i++) 33 { 34 cin >> Dist[i]; 35 Sum += Dist[i]; 36 } 37 cin >> M; 38 int v1, v2; 39 for (int i = 0; i < M; i++) 40 { 41 cin >> v1 >> v2; 42 Ans(v1, v2); 43 } 44 }
ac代码
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <climits> 3 #include<iostream> 4 #include<vector> 5 #include<queue> 6 #include<map> 7 #include<set> 8 #include<stack> 9 #include<algorithm> 10 #include<string> 11 #include<cmath> 12 using namespace std; 13 vector<int>Dist; 14 int sum; 15 int main() 16 { 17 int N; 18 cin >> N; 19 Dist.resize(N + 1); 20 for (int i = 1; i <=N; i++) 21 { 22 int temp; 23 cin >> temp; 24 sum += temp; 25 Dist[i] = sum; 26 } 27 int M; 28 cin >> M; 29 int v1, v2; 30 for (int i = 0; i < M; i++) 31 { 32 cin >> v1 >> v2; 33 if (v1 > v2) 34 swap(v1,v2); 35 int s1 = Dist[v2 - 1] - Dist[v1 - 1]; 36 int s2 = sum - s1; 37 if (s1 > s2) 38 cout << s2 << endl; 39 else 40 cout << s1 << endl; 41 } 42 }
标签：names ase really secure follow limit each res close
原文地址：https://www.cnblogs.com/57one/p/12037128.html
