A. Arrays------(Codeforces Round #317 [AimFund Thanks-Round] (Div. 2))
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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.
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.
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).
3 32 11 2 33 4 5
YES
3 33 31 2 33 4 5
NO
5 23 11 1 1 1 12 2
YES
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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