A. Arrays
time limit per test
memory limit per test
input
output
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
Input
nA, nB (1 ≤ nA, nB ≤ 105), separated by a space — the sizes of arrays A and B, correspondingly.
k and m (1 ≤ k ≤ nA, 1 ≤ m ≤ nB), separated by a space.
nA numbers a1, a2, ... anA (9 ≤ a1 ≤ a2 ≤ ... ≤ anA ≤ 109), separated by spaces — elements of array A.
nB integers b1, b2, ... bnB (9 ≤ b1 ≤ b2 ≤ ... ≤ bnB ≤ 109), separated by spaces — elements of array B.
Output
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
A and number 3 from array B
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:
.
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
int a[100100],b[100010];
int n,m;
int k,t;
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
scanf("%d%d",&k,&t);
for(int i=0; i<n; i++)
{
scanf("%d",&a[i]);
}
for(int i=0; i<m; i++)
{
scanf("%d",&b[i]);
}
if(a[k-1] < b[m-t])
{
printf("YES\n");
}
else
{
printf("NO\n");
}
}
return 0;
}
