Ants
Posted luckykid
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Description
An army of ants walk on a horizontal pole of length l cm, each with a constant speed of 1 cm/s. When a walking ant reaches an end of the pole, it immediatelly falls off it. When two ants meet they turn back and start walking in opposite directions. We know the original positions of ants on the pole, unfortunately, we do not know the directions in which the ants are walking. Your task is to compute the earliest and the latest possible times needed for all ants to fall off the pole.
Input
Output
Sample Input
2
10 3
2 6 7
214 7
11 12 7 13 176 23 191
Sample Output
4 8
38 207
Source
假设每一只蚂蚁都不回头直接继续而行,因此最大的距离就是蚂蚁距离两边距离的最大值,最小值就是距离两边的距离的最小值中的最大值
1 #include<iostream> 2 #include<algorithm> 3 using namespace std; 4 int main() { 5 int n; 6 cin >> n; 7 while (n--) { 8 int length, antNumber; 9 int mx = 0, leftLength, mn = 0; 10 cin >> length >> antNumber; 11 while (antNumber--) { 12 //求最大距离 13 cin >> leftLength; 14 mx = max(max(length - leftLength, leftLength), mx); 15 //求最小距离 16 mn = max(min(length - leftLength, leftLength), mn); 17 } 18 cout << mn << " " << mx << endl; 19 } 20 return 0; 21 }
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