Acdream 1099 calculates the k-th largest

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Yaoyao's k-th time limit: 10000/5000 ms (Java/others) memory limit: 512000/256000 KB (Java/others) submitstatisticnext problemproblem description

One day, Yao (tsyao), the cute girl, was bored and came to play with you. But you don't know what to play... After a while, the witty Yaoyao proposed: "Well, you saidNIntegersXIAnd then say a number at will.K, I can quickly express the numbersKA large number ."

Input

Two integers in row 1stN, KSeparated by spaces;

2nd rows haveNIntegers (the same number can appear, all are randomly generated), also separated by spaces.

0 <n ≤ 5*10 ^ 6, 0 <k ≤ n

1 ≤ Xi ≤ 10 ^ 8

Output KA large number. Sample Input

5 25 4 1 3 1

Sample output

4

In hint numbers such as 2, 2, and 1, the first digit is 2, the second digit is 2, and the third digit is 1.

Because N is too large, you need to use input hanging, and then quickly select O (n. 1a Accepted code:

1 /************************************** * *********************************** 2> File Name: kth. CPP 3> author: stomach_ache 4> mail: [email protected] 5> created time:, Saturday, August 02, 2014> propose: acdream 7 ************************************** * **********************************/8 // enter + quick selection of 9 # include <cmath> 10 # include <string> 11 # include <cstdio> 12 # include <fstream> 13 # include <CS Tring> 14 # include <iostream> 15 # include <algorithm> 16 using namespace STD; 17 18 int N, K; 19 int A [1, 5000002]; 20 21 int read () {22 int x = 0; 23 char CH = ''; 24 while (CH <'0' | ch> '9') CH = getchar (); 25 while (CH> = '0' & Ch <= '9') x = x * 10 + CH-'0', CH = getchar (); 26 return X; 27} 28 29 int sort (int l, int R) {30 if (L> = r) return a [l]; 31 int substring = A [(L + r) /2]; 32 int I = L, j = r; 33 (;) {34 while (I <J & A [I] <= cursor) I ++; 35 while (I <J & A [J]> = callback) j --; 36 if (I <j) Swap (A [I], a [J]); 37 Else break; 38 I ++; j --; 39} 40 swap (A [I], a [(L + r)/2]); 41 if (I = k) return a [I]; 42 if (I <k) return sort (I + 1, R); 43 else return sort (L, i-1); 44} 45 46 int main (void) {47 While (~ Scanf ("% d", & N, & K) {48 for (INT I = 1; I <= N; I ++) 49 A [I] = read (); 50 k = n-k + 1; 51 int ans = sort (1, N); 52 printf ("% d \ n ", ans); 53} 54 return 0; 55}