SWUST OJ Gold Nuggets Distribution(0490)

Gold Nuggets Distribution(0490)

 

Time limit(ms): 1000 Memory limit(kb): 65535 Submission: 421 Accepted: 234

 

Description

Bosses have a bag of gold. Each month there will be two employees to their outstanding performance were a gold award. In line with the rituals, ranking first employees will receive the heaviest nugget, ranked second employee will be the lightest gold. Under this method, unless the Nuggets to a new bag, otherwise employees received the first gold always be employees than the second-Nuggets. If there is a new cyclical Nuggets joined bag, every month to find the lightest and most of the Nuggets. Assuming that there is a comparison of weight machines, we hope that at least by comparison to identify the number and the lightest weight of gold.

(题目描述就不翻译了，简而言之的意思就是，找出数组中的最大元素和最小元素，有个要求就是，比较的次数最少)

 

Input

two lines 
the first line is a number of gold nuggets and less than 50000 
the second line are weight of every gold nugget.

 

Output

two numbers 
the first is weight of the heaviest nugget 
the second is weight of the lightest gold

 

Sample Input

6

5 2 7 6 3 4

 

Sample Output

7 2

 

 Hint

Source

MrYang

 

#include<iostream>
using namespace std;

int a[2000000];

int main()
{
    int n,i,j,max=0,min=99999;
    cin>>n;
    for(i=0;i<n;i++)
        cin>>a[i];
    for(i=0,j=n-1;i<=j;i++,j--)
    {
        if(a[i]<a[j])
        {
            if(a[i]<min)
                min=a[i];
            if(a[j]>max)
                max=a[j];
        }
        else
        {
            if(a[j]<min)
                min=a[j];
            if(a[i]>max)
                max=a[i];
        }
    }
    cout<<max<<' '<<min<<endl;
    return 0;
}

 

注：两个数比较，则那个大的数，不可能成为这组数中最小的数，同理，那个小的数，不可能成为这组数中最大的数。从常用的 2*n 次比较，变成了（n/2）*3=1.5*n 次比较。