UVA 10714 Ants
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The first line of input contains one integer giving the number of cases that follow. The data for each case start with two integer numbers: the length of the pole (in cm) and n, the number of ants residing on the pole. These two numbers are followed by n integers giving the position of each ant on the pole as the distance measured from the left end of the pole, in no particular order. All input integers are not bigger than 1000000 and they are separated by whitespace.
For each case of input, output two numbers separated by a single space. The first number is the earliest possible time when all ants fall off the pole (if the directions of their walks are chosen appropriately) and the second number is the latest possible such time.
Sample input
210 32 6 7214 711 12 7 13 176 23 191
Output for sample input
4 838 207
所有蚂蚁在一条线上,方向未知,问最后一只蚂蚁离开线段的时间最短和最长分别是多少。
两只蚂蚁相遇后会转向回去,其实就相当于没有相遇,所以只要考虑最远和最近的就好了。
#include<stdio.h>#include<iostream> #include<algorithm>#include<cmath>#include<cstdio>#include<string>#include<cstring>using namespace std;int n, t, maxn, minn, m, a;int main(){cin >> t;while (t--){cin >> n >> m;maxn = 0;minn = 0;while (m--){scanf("%d", &a);maxn = max(maxn, max(a, n - a));minn = max(minn, min(a, n - a));}printf("%d %d\n", minn, maxn);}return 0;}
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