Codeforces Round #317 [AimFund Thanks-Round] (Div. 2) A. Arrays | 书脊
http://codeforces.com/contest/572/problem/A
题目大意: 给两个数组, 问是否能在第一个数组找到k个数,在第二个数组找到m个数, 使得所有在第一个数组找到的数小于第二个数组找到的数. 数组已排序.
分析: 因为数组已经排序, 所以如果在第一个数组前k个数字小于第二个数组后m个数字, 那么打印YES.如果不是, 打印NO
http://codeforces.com/contest/572/problem/A
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 arrayA.
The fourth line contains nB integers b1, b2, ... bnB ( - 109 ≤ b1 ≤ b2 ≤ ... ≤ bnB ≤ 109), separated by spaces — elements of arrayB.
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).
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: 因为数组已经排序, 所以如果在第一个数组前k个数字小于第二个数组后m个数字, 那么打印YES.如果不是, 打印NO
public void solve(int testNumber, InputReader in, OutputWriter out) {
int na = in.readInt();
int nb = in.readInt();
int k = in.readInt();
int m = in.readInt();
int[] naAry = IOUtils.readIntArray(in, na);
int[] nbAry = IOUtils.readIntArray(in, nb);
if (naAry[k - 1] >= nbAry[nb - m])
out.print("NO");
else
out.print("YES");
}
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