A. Arrays------(Codeforces Round #317 [AimFund Thanks-Round] (Div. 2))

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A. Arrays
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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 32 11 2 33 4 5
output
YES
input
3 33 31 2 33 4 5
output
NO
input
5 23 11 1 1 1 12 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.


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;    }}


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