Codeforces 572 A. Arrays

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                                        ***A. Arrays***

You are given two arrays A and B consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose k numbers 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 array A was strictly less than any number chosen in array B. Otherwise, print "NO" (without the quotes).

题目大意：给你连个数组 a 和 b ，两个数 m, k 分别是数组 a 的 m 个数和数组 b 的 k 个数，然后是数组a和b的值，看是否满足数组 b 中的 k 个数全比数组 a 的 m 个数大（a 和 b已经从小到大排好序）

解题思路：因为a 和 b已经排好序，所以只需要找 a 中的第 m 个数和 b
中的倒数第 k 个数就行 ，换句话说就是，找 a 中最大的数，和b中最小的数

上代码：

/*
Date : 2015-8-27 晚上
Author : ITAK

Motto :

今日的我要超越昨日的我，明日的我要胜过今日的我；
以创作出更好的代码为目标，不断地超越自己。
*/
#include <iostream>
#include <cstdio>
using namespace std;

const int maxn = 1e5 + 5;

int a[maxn], b[maxn];
int main()
{
    int na, nb, k, m;
    while(cin>>na>>nb)
    {
        cin>>k>>m;
        for(int i=0; i<na; i++)
            cin>>a[i];
        for(int i=0; i<nb; i++)
            cin>>b[i];
        if(a[k-1] < b[nb-m])
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}
/**
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
**/