最多时间就是每只蚂蚁选择最久的爬行方式
最少时间就是每只蚂蚁选择最快地爬行方式
#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
int a,b,i,T,len,n,t;
cin>>T;
while(T--)
{
cin>>len>>n;
a=b=0;
while(n--)
{
cin>>t;
a=max(a,min(len-t,t));
b=max(b,max(len-t,t));
}
cout<<a<<" "<<b<<endl;
}
return 0;
}
| 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) andn, the number of ants residing on the pole. These two numbers are followed byn 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
原文：http://blog.csdn.net/stl112514/article/details/40459563