http://poj.org/problem?id=2833
The AverageTime Limit: 6000MS Memory Limit: 10000K
Total Submissions: 7443 Accepted: 2288
Case Time Limit: 4000MS
Description
In a speech contest, when a contestant finishes his speech, the judges will then grade his performance. The staff remove the highest grade and the lowest grade and compute the average of the rest as the contestant’s final grade. This is an easy problem because usually there are only several judges.
Let’s consider a generalized form of the problem above. Given n positive integers, remove the greatest n1 ones and the least n2 ones, and compute the average of the rest.
Input
The input consists of several test cases. Each test case consists two lines. The first line contains three integers n1, n2 and n (1 ≤ n1, n2 ≤ 10, n1 + n2 < n ≤ 5,000,000) separate by a single space. The second line contains n positive integers ai (1 ≤ ai ≤ 108 for all i s.t. 1 ≤ i ≤ n) separated by a single space. The last test case is followed by three zeroes.
Output
For each test case, output the average rounded to six digits after decimal point in a separate line.
Sample Input
1 2 5
1 2 3 4 5
4 2 10
2121187 902 485 531 843 582 652 926 220 155
0 0 0
Sample Output
3.500000
562.500000
Hint
This problem has very large input data. scanf and printf are recommended for C++ I/O.
The memory limit might not allow you to store everything in the memory.
題目限制了記憶體。但是題目給的n1和n2很小
是以用優先隊列或堆來維護2個最大和最小值的數組
用了3個方法，堆的效率最高。
1.直接模拟
//Memory: 160K Time: 3766MS
#include <cstdio>
#include <algorithm>
using namespace std;
__int64 max_t[11],min_t[11],total,data;
int main()
{
int i,n1,n2,n;
double ave;
while(1)
{
memset(max_t,0,sizeof(max_t));
memset(min_t,100000001,sizeof(min_t));//初始化數組
total = 0;
scanf("%d%d%d",&n1,&n2,&n);
if(n1+n2+n == 0)
break;
for(i = 0;i < n;i ++)
{
scanf("%I64d",&data);
total = total+data;//求總和
sort(max_t,max_t+n1);//對存儲最大值的數組排序
sort(min_t,min_t+n2);//對存儲最小值的數組排序
if(data > max_t[0])//與最大值數組中的最小值比較
max_t[0] = data;
if(min_t[n2-1] > data)//與最小值數組中的最大值比較
min_t[n2-1] = data;
}
for(i = 0;i < n1;i ++)
total = total - max_t[i];
for(i = 0;i < n2;i ++)
total = total - min_t[i];
ave = total/(double)(n-n1-n2);//求平均值
printf("%.6lf/n",ave);
}
return 0 ;
}
2.隊列
//Memory: 176K Time: 3297MS
#include <iostream>
#include <queue>
using namespace std;
int Max[15],Min[15];
int s1,s2;
int cmp(int a,int b)
{
return a>b;
}
int main()
{
int n1,n2,n,i,x;
__int64 sum;
while (scanf("%d%d%d",&n1,&n2,&n)&&(n1+n2+n))
{
priority_queue <int>q1; //從大到小。存小數
priority_queue <int,vector<int>,greater<int> >q2;
n1++,n2++;
sum=0;
for (i=0;i<n;i++)
{
scanf("%d",&x);
sum+=x;
q2.push(x);
q1.push(x);
if(q1.size()>n2)
q1.pop();
if(q2.size()>n1)
q2.pop();
}
q1.pop();
q2.pop();
while(!q1.empty())
{
sum-=q1.top();
q1.pop();
}
while(!q2.empty())
{
sum-=q2.top();
q2.pop();
}
n-=(n1+n2-2);
printf("%.6lf/n",sum/(n+0.0));
}
return 0;
}
3.堆
用法 ：http://blog.csdn.net/zsc09_leaf/archive/2011/04/10/6313716.aspx
//Memory: 160K Time: 2579MS
#include <iostream>
#include <algorithm>
using namespace std;
int Max[15],Min[15];
int s1,s2;
int cmp(int a,int b)
{
return a>b;
}
int main()
{
int n1,n2,n,i,x;
double sum;
while (scanf("%d%d%d",&n1,&n2,&n)&&(n1+n2+n))
{
sum=0;
for (i=0;i<11;i++)
{
Max[i]=0;
Min[i]=2000000000;
}
for(i=0;i<n;i++)
{
scanf("%d",&x);
sum+=x;
if(x>Max[n1])
{
Max[n1]=x;
make_heap(Max,Max+n1+1,cmp);
pop_heap(Max,Max+n1+1,cmp);
}
if(x<Min[n2])
{
Min[n2]=x;
make_heap(Min,Min+n2+1);
pop_heap(Min,Min+n2+1);
}
}
for(i=0;i<n1;i++)
sum-=Max[i];
for (i=0;i<n2;i++)
sum-=Min[i];
n-=(n1+n2);
printf("%.6lf/n",sum/(n+0.0));
}
return 0;
}