A. Arrays------（Codeforces Round #317 [AimFund Thanks-Round] (Div. 2)）_first line contains two integers na, nb (1 ≤ na, n-优快云博客

本文介绍了一个简单的算法问题，即判断是否可以从两个已排序的整数数组A和B中分别选择k个和m个元素，使得所有从A中选择的元素都严格小于从B中选择的所有元素。通过输入两个数组及k和m的值，使用快速排序方法，可以有效地解决这一问题。

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You are given two arrays A and B consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose knumbers in array A and choose m numbers in array B so that any number chosen in the first array is strictly less than any number chosen in the second array.

Input

The first line contains two integers nA, nB (1 ≤ nA, nB ≤ 105), separated by a space — the sizes of arrays A and B, correspondingly.

The second line contains two integers k and m (1 ≤ k ≤ nA, 1 ≤ m ≤ nB), separated by a space.

The third line contains nA numbers a1, a2, ... anA ( - 109 ≤ a1 ≤ a2 ≤ ... ≤ anA ≤ 109), separated by spaces — elements of array A.

The fourth line contains nB integers b1, b2, ... bnB ( - 109 ≤ b1 ≤ b2 ≤ ... ≤ bnB ≤ 109), separated by spaces — elements of array B.

Output

Print "YES" (without the quotes), if you can choose k numbers in array A and m numbers in array B so that any number chosen in arrayA was strictly less than any number chosen in array B. Otherwise, print "NO" (without the quotes).

Sample test(s)

input
3 3
2 1
1 2 3
3 4 5

output
YES

input
3 3
3 3
1 2 3
3 4 5

output
NO

input
5 2
3 1
1 1 1 1 1
2 2

output
YES

Note

In the first sample test you can, for example, choose numbers 1 and 2 from array A and number 3 from array B (1 < 3 and 2 < 3).

In the second sample test the only way to choose k elements in the first array and m elements in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in A will be less than all the numbers chosen in B: $\text{[math]}$ .

分析：水题。PS:感觉还是应该先做A题，10分钟左右的事，再做第二题不迟。

CODE：

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int na,nb;
    while(cin>>na>>nb){
        int k,m,a[100005],b[100005];
        cin>>k>>m;
        for(int i=0;i<na;i++)
            cin>>a[i];
        for(int i=0;i<nb;i++)
            cin>>b[i];
        sort(a,a+na);
        sort(b,b+nb);
        if(a[k-1]<b[nb-m])
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
}