2013
12-03

# Who’s in the Middle

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.

* Line 1: A single integer N

* Lines 2..N+1: Each line contains a single integer that is the milk output of one cow.

* Line 1: A single integer that is the median milk output.

5
2
4
1
3
5

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.


1)先排序,然后去中间哪一个即可

/*
* 1157_1.cpp
*
*  Created on: 2013年8月10日
*/

#include <iostream>

using namespace std;

int main(){
int n;
while(scanf("%d",&n)!=EOF){
int i,o[n];
for(i = 0 ; i < n ; ++i){
scanf("%d",&o[i]);
}

sort(o,o+n);

cout<<o[n/2]<<endl;
}
}

1. 为什么for循环找到的i一定是素数叻，而且约数定理说的是n=p1^a1*p2^a2*p3^a3*…*pk^ak，而你每次取余都用的是原来的m，也就是n

2. 我还有个问题想请教一下，就是感觉对于新手来说，递归理解起来有些困难，不知有没有什么好的方法或者什么好的建议？