UVA - 10714
| Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
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.
2 10 3 2 6 7 214 7 11 12 7 13 176 23 191
4 8 38 207
Source
思路：只需要遍历所有蚂蚁，把每个蚂蚁走出木杆的最长时间，最短时间分别求出来，取其中的最大值，就是题目所问的所有蚂蚁离开木杆的最短时间和最长时间
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#define INF 0x3fffffff
using namespace std;
int main() {
int T;
scanf("%d", &T);
while(T--) {
int len, n, MA = 0, MI = 0;
scanf("%d %d", &len, &n);
for(int i = 0; i < n; i++) {
int t;
scanf("%d", &t);
int ma = len - t;
int mi = len - ma;
if(ma < mi) swap(ma, mi);
MA = max(MA, ma);
MI = max(MI, mi);
}
printf("%d %d\n", MI, MA);
}
return 0;
}
原文地址：http://blog.csdn.net/u014355480/article/details/44500033
