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

2015-08-26 18:05 791 查看
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 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:

.

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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