codeforces 572A Arrays

本文介绍了一个简单的算法问题,即如何从两个已排序的整数数组中分别选择k个和m个元素,使得第一个数组中的所有选中元素严格小于第二个数组中的所有选中元素。通过实例演示了如何快速解决这个问题。

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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 array Awas strictly less than any number chosen in array B. Otherwise, print "NO" (without the quotes).

Examples
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.


题意自己翻译,水题,取a数组k个,取最小的,b数组取m个,取最大的,直接比较新数组a1的最大值和新数组b1的最小值,这两个值的大小关系。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5;
int main()
{
    int na,nb;
    int k,m;
    int a[N+5],b[N+5];
    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];
    sort(a,a+na);
    sort(b,b+nb);
    reverse(b,b+nb);
    if(a[k-1]<b[m-1])
        cout<<"YES"<<endl;
    else
        cout<<"NO"<<endl;
    return 0;
}


### Codeforces Round 1001 Problems and Solutions #### Problem A: Contest Timing Calculation To determine the minimum moment of time when Pasha can submit all solutions during a contest, one must consider both preparation times for each problem as well as submission delays. For instance, if there are multiple problems with varying solving durations \(a_i\) and submission waiting periods \(b_i\), then calculating the optimal sequence becomes crucial[^1]. Example Input: ```plaintext 3 2 3 1 0 ``` Output would be calculated based on these parameters ensuring minimal total elapsed time. #### Problem B: Test Case Generation Given an integer \(n\) where \(1 \leq n \leq 100\), this section focuses on generating appropriate pairs (\(a_i\),\(b_i\)) that satisfy specific constraints related to test cases generation. Each pair represents inputs leading to valid outputs under certain conditions defined within K1o0n's algorithm implementation scenario described earlier[^2]. An example set might look like: Input: ```plaintext 3 2 3 1 0 ``` Expected Output: ```plaintext 3 20 18 219 216 2218 2214 ``` This demonstrates how different values affect outcome correctness according to predefined ruleset provided initially. #### Problem E: Array Manipulation Challenge Involves manipulating arrays through deletion operations while adhering strictly to memory limits (up to 256 MB) over short execution windows (~2 seconds). Given an initial list `a` containing up to \(n\) elements alongside another value `x`, tasks revolve around strategically removing segments from original data structure without violating imposed restrictions regarding element bounds or overall performance metrics outlined previously[^4]. Sample code snippet demonstrating approach could involve iterating backwards across indices checking against criteria before executing removals efficiently. ```python def range_deleting(a, x): result = [] temp_stack = [] for num in reversed(a): if not temp_stack or any(num <= item <= x for item in temp_stack[-min(x, len(temp_stack)):]): temp_stack.append(num) else: result.extend(reversed(temp_stack)) temp_stack.clear() temp_stack.append(num) result.extend(reversed(temp_stack)) return result[::-1] ``` --related questions-- 1. How does adjusting variable ranges impact computational complexity? 2. What strategies optimize runtime efficiency during competitive programming contests? 3. Can alternative algorithms achieve better results compared to those presented here? 4. Are there scenarios where increasing available resources improves solution feasibility significantly?
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