Ants poj 1852

Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 14924 Accepted: 6487 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 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. Output 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 2 10 3 2 6 7 214 7 11 12 7 13 176 23 191 Sample Output 4 8

38 207

题意：给你一根棍子的长度，你只知道蚂蚁距离棍子左端多长，两个蚂蚁相碰就各种掉头，问你最短最长距离。

思路：很简单，当两个蚂蚁掉头的时候仍然可以认为他们在继续走。。

#include<stdio.h> #define MAX_N 1000000+5 int L,n; int x[MAX_N]; int max(int x, int y) { if(x>y) return x; return y; } int min(int x,int y) { if(x<y) return x; return y; } int main() { int a; scanf("%d",&a); while(a--) { scanf("%d%d",&L,&n); for(int i=0;i<n;i++) scanf("%d",&x[i]); int minT=0; for(int i=0;i<n;i++) { minT=max(minT,min(x[i],L-x[i])); } int maxT = 0; for(int i = 0; i < n;i++) { maxT=max(maxT,max(x[i],L-x[i])); } printf("%d %d\n",minT,maxT); } return 0; }