改变 k 个元素，使得(a1^2+a2^2+…+an^2)<=(a1+a2+…+an)变为真
原文:https://www . geesforgeks . org/change-k-elements-so-so-a12-a22-an2/
给定一个大小为 n 的数组 Arr ，任务是判断是否有可能将这个序列中最多 K 个元素改变为任意正整数，使得以下条件成立。
示例:
Input:N = 2, Arr[] = {1, 2}, K = 2
Output: Possible
(As A[2] can be change to 1)
Input: N = 2, Arr[] = {5, 6}, K = 1
Output: Not Possible
(As we can only change 1 element to any arbitrary number
and after changing it doesn't satisfy above condition)
方法:当数组的所有元素都等于 1 时，那么只能满足给定的方程，否则不能满足。
- 遍历数组并计算 1 的个数。
- 如果 K >=(数组大小，即 N-计数)，则返回真，否则返回假。
下面是上述方法的实现:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Function that will tell
// whether it is possible or Not
int Series(int Arr[], int N, int K)
{
int count = 0;
for (int i = 0; i < N; i++)
if (Arr[i] == 1)
count++;
if (K >= (N - count))
return 1;
else
return 0;
}
// Driver code
int main()
{
int Arr[] = { 5, 1, 2 };
int N = sizeof(Arr) / sizeof(Arr[0]);
int K = 2;
// Calling function.
int result = Series(Arr, N, K);
if (result == 1)
cout << "Possible";
else
cout << "Not Possible";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
//Java implementation of above approach
import java.io.*;
class GFG {
// Function that will tell
// whether it is possible or Not
static int Series(int Arr[], int N, int K)
{
int count = 0;
for (int i = 0; i < N; i++)
if (Arr[i] == 1)
count++;
if (K >= (N - count))
return 1;
else
return 0;
}
// Driver code
public static void main (String[] args) {
int Arr[] = { 5, 1, 2 };
int N = Arr.length;
int K = 2;
// Calling function.
int result = Series(Arr, N, K);
if (result == 1)
System.out.println ("Possible");
else
System.out.println( "Not Possible");
}
//This Code is Contributed by ajit
}
Python 3
# Python implementation of
# above approach
# Function that will tell
# whether it is possible or Not
def Series(Arr, N, K):
count = 0
for i in range(N):
if Arr[i] == 1:
count += 1
if K >= (N - count):
return 1
return 0
# Driver code
Arr = [5, 1, 2]
N = len(Arr)
K = 2
result = Series(Arr, N, K)
if result == 1:
print("Possible")
else:
print("Not Possible")
# This code is contributed
# by Shrikant13
C
//C# implementation of above approach
using System;
public class GFG{
// Function that will tell
// whether it is possible or Not
static int Series(int []Arr, int N, int K)
{
int count = 0;
for (int i = 0; i < N; i++)
if (Arr[i] == 1)
count++;
if (K >= (N - count))
return 1;
else
return 0;
}
// Driver code
static public void Main (){
int []Arr = { 5, 1, 2 };
int N = Arr.Length;
int K = 2;
// Calling function.
int result = Series(Arr, N, K);
if (result == 1)
Console.WriteLine ("Possible");
else
Console.WriteLine( "Not Possible");
}
//This Code is Contributed by akt_mit
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of above approach
// Function that will tell
// whether it is possible or Not
function Series($Arr, $N, $K)
{
$count = 0;
for ($i = 0; $i < $N; $i++)
if ($Arr[$i] == 1)
$count++;
if ($K >= ($N - $count))
return 1;
else
return 0;
}
// Driver code
$Arr = array( 5, 1, 2 );
$N = sizeof($Arr);
$K = 2;
// Calling function.
$result = Series($Arr, $N, $K);
if ($result == 1)
echo "Possible";
else
echo "Not Possible";
// This code is contributed
// by Sach_Code
?>
java 描述语言
<script>
// Javascript implementation of above approach
// Function that will tell
// whether it is possible or Not
function Series(Arr, N, K)
{
var count = 0;
for(var i = 0; i < N; i++)
if (Arr[i] == 1)
count++;
if (K >= (N - count))
return 1;
else
return 0;
}
// Driver code
var Arr = [ 5, 1, 2 ];
var N = Arr.length;
var K = 2;
// Calling function.
var result = Series(Arr, N, K);
if (result == 1)
document.write("Possible");
else
document.write("Not Possible");
// This code is contributed by rutvik_56
</script>
Output:
Possible
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