Poj 2388 Who's in the Middle
2014-12-04 13:07
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1.Link:
http://poj.org/problem?id=2388
2.Content:
Who's in the Middle
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.
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
Sample Output
Hint
INPUT DETAILS:
Five cows with milk outputs of 1..5
OUTPUT DETAILS:
1 and 2 are below 3; 4 and 5 are above 3.
Source
USACO 2004 November
3.Method:
4.Code:
5.Reference:
http://poj.org/problem?id=2388
2.Content:
Who's in the Middle
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 31854 | Accepted: 18541 |
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
* 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
Sample Output
3
Hint
INPUT DETAILS:
Five cows with milk outputs of 1..5
OUTPUT DETAILS:
1 and 2 are below 3; 4 and 5 are above 3.
Source
USACO 2004 November
3.Method:
4.Code:
1 #include "iostream" 2 #include "stdlib.h" 3 #include "stdio.h" 4 #include <time.h> 5 #define N 10002 6 using namespace std; 7 int a ; 8 int Partition(int a[],int p,int r) 9 { 10 int tmp = a[p]; 11 while(p<r) 12 { 13 while(p<r && a[r]>=tmp) r--; 14 a[p]=a[r]; 15 while(p<r && a[p]<=tmp) p++; 16 a[r]=a[p]; 17 } 18 a[p]=tmp; 19 return p; 20 } 21 int RandomizePartition(int a[],int p,int r) 22 { 23 int i = rand()%(r-p+1)+p; 24 int tmp=a[i]; 25 a[i]=a[p]; 26 a[p]=tmp; 27 return Partition(a,p,r); 28 } 29 30 int RandomizedSelect(int a[],int p,int r,int k) 31 { 32 if(p==r) return a[p]; 33 int i = RandomizePartition(a,p,r),j = i-p+1; 34 if(k<=j) return RandomizedSelect(a,p,i,k); 35 else return RandomizedSelect(a,i+1,r,k-j); 36 } 37 int main() 38 { 39 srand(time(NULL));//初始化随机数生成器 40 int n; 41 int i; 42 scanf("%d",&n); 43 for(i=0;i<n;i++) scanf("%d",&a[i]); 44 //qsort(a,0,n-1); 45 //printf("%d\n",a[n/2]); 46 printf("%d\n",RandomizedSelect(a,0,n-1,(n+1)/2)); 47 system("pause"); 48 4000 return 0; 49 }
5.Reference:
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