Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 42058 | Accepted: 24335 |
Description
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
* Lines 2..N+1: Each line contains a single integer that is the milk output of one cow.
Output
Sample Input
5 2 4 1 3 5
Sample Output
3
//下面是代碼,題意就是排好序後求中間的那個數
#include<stdio.h>
#define MAX 10005
int a[MAX];
void quick_sort(int s[], int l, int r)
{
if (l < r)
{
int i = l, j = r, x = s[l];
while (i < j)
{
while(i < j && s[j] >= x) // 從右向左找第一個小於x的數
j--;
if(i < j)
s[i++] = s[j];
while(i < j && s[i] < x) // 從左向右找第一個大於等於x的數
i++;
if(i < j)
s[j--] = s[i];
}
s[i] = x;
quick_sort(s, l, i - 1); // 遞歸調用
quick_sort(s, i + 1, r);
}
}
int main()
{
int T;
scanf("%d",&T);
for(int i=0;i<T;i++)
{
scanf("%d",&a[i]);
}
quick_sort(a,0,T-1);
printf("%d\n",a[T/2]);
return 0;
}