Description
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. 输入N个数,输出升序排列后中间那个数.
(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. 输入N个数,输出升序排列后中间那个数.
Input
* Line 1: A single integer N * Lines 2..N+1: Each line contains a single integer that is the milk output of one cow.
Output
* Line 1: A single integer that is the median milk output.
Sample Input
5
2
4
1
3
5
INPUT DETAILS:
Five cows with milk outputs of 1..5
2
4
1
3
5
INPUT DETAILS:
Five cows with milk outputs of 1..5
Sample Output
3
OUTPUT DETAILS:
1 and 2 are below 3; 4 and 5 are above 3.
OUTPUT DETAILS:
1 and 2 are below 3; 4 and 5 are above 3.
金组题就是输出中位数……这题目是不是出的有问题……
#include<cstdio>
#include<algorithm>
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n,a[10010];
int main()
{
n=read();
for (int i=1;i<=n;i++)a[i]=read();
sort(a+1,a+n+1);
printf("%d",a[(1+n)>>1]);
return 0;
}