标签：highlight tput details div most main input amp mes
FJ is surveying his herd to find the most average cow. He wants to know how much milk this ‘median‘ cow gives: half of the cows give as much or more than the median; half give as much or less.
Given an odd number of cows N (1 <= N < 10,000) and their milk output (1..1,000,000), find the median amount of milk given such that at least half the cows give the same amount of milk or more and at least half give the same or less.
Input
Output
5 2 4 1 3 5Sample Output
3Hint
#include <iostream>
using namespace std;
int main( )
{
int num;
int arr[1001];
while(cin>>num)
{
int i,temp,counter=0;
for(i=0;i<1001;i++)
arr[i]=0;
for(i=0;i<num;i++)
{
cin>>temp;
arr[temp]++;
}
for(i=1;i<=1000;i++)
{
for(temp=1;temp<=1000;temp++)
{
if(arr[temp]!=0)
{
arr[temp]--;
break;
}
}
for(temp++;temp<=1000;temp++)
{
if(arr[temp]!=0)
{
arr[temp]--;
counter++;
}
}
}
cout<<counter<<endl;
}
return 0;
}
标签：highlight tput details div most main input amp mes
原文地址：http://www.cnblogs.com/wongyi/p/7773902.html
