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
1 #include<iostream>
2 using namespace std;
3
4 int a[2000000];
5
6 int main()
7 {
8 int n,i,j,max=0,min=99999;
9 cin>>n;
10 for(i=0;i<n;i++)
11 cin>>a[i];
12 for(i=0,j=n-1;i<=j;i++,j--)
13 {
14 if(a[i]<a[j])
15 {
16 if(a[i]<min)
17 min=a[i];
18 if(a[j]>max)
19 max=a[j];
20 }
21 else
22 {
23 if(a[j]<min)
24 min=a[j];
25 if(a[i]>max)
26 max=a[i];
27 }
28 }
29 cout<<max<<‘ ‘<<min<<endl;
30 return 0;
31 }
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。