ACdream OJ 1099 瑶瑶的第K大 --分治+IO优化

/*
* this code is made by whatbeg
* Problem: 1099
* Verdict: Accepted
* Submission Date: 2014-06-15 00:13:53
* Time: 4340 MS
* Memory: 21212 KB
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <iomanip>
using namespace std;
#define N 1007

//复杂度O(n)
int Max_K(int a[],int low,int high,int k)
{
if(k <= 0 || k > high-low+1)
return -1;
int flag = low + abs(rand())%(high-low+1);  //随机选择一个基准点
swap(a[low],a[flag]);
int mid = low;
int cnt = 1;
for(int i=low+1;i<=high;i++)
{
if(a[i] > a[low])   //遍历，把较大的数放在数组左边
{
swap(a[++mid],a[i]);
cnt++;
}
}
//比基准点大的数的个数为cnt-1
swap(a[mid],a[low]);    //将基准点放在左、右两部分的分界处
if(cnt > k)
return Max_K(a,low,mid-1,k);
else if(cnt < k)
return Max_K(a,mid+1,high,k-cnt);
else
return mid;
}

int Min_K(int a[],int low,int high,int k)
{
if(k <= 0 || k > high-low+1)
return -1;
int flag = low + abs(rand())%(high-low+1);
swap(a[low],a[flag]);
int mid = low;
int cnt = 1;
for(int i=low+1;i<=high;i++)
{
if(a[i] < a[low])
{
swap(a[++mid],a[i]);
cnt++;
}
}
swap(a[mid],a[low]);
if(k < cnt)
return Min_K(a,low,mid-1,k);
else if(k > cnt)
return Min_K(a,mid+1,high,k);
else
return mid;
}

inline int in()
{
char ch;
int a = 0;
while((ch = getchar()) == ' ' || ch == '\n');
a += ch - '0';
while((ch = getchar()) != ' ' && ch != '\n')
{
a *= 10;
a += ch - '0';
}
return a;
}

inline void out(int a)
{
if(a >= 10)
out(a / 10);
putchar(a % 10 + '0');
}

int a[5000006];

int main()
{
int n,k;
while(scanf("%d%d",&n,&k)!=EOF)
{
getchar();
for(int i=0;i<n;i++)
a[i] = in();
int res = Max_K(a,0,n-1,k);
out(a[res]);
puts("");
}
return 0;
}