给定两个整数N和K ,任务是找到K个数字( A 1 ,A 2 ,…,A K ),使得∑ i = 1 K A i等于N且∑ i = 1 K A i 2最大。
例子:
Input: N = 3, K = 2
Output: 1 2
Explanation: The two numbers are 1 and 2 as their sum is equal to N and 12 + 22 is maximum.
Input: N = 10, K = 3
Output: 1 8 1
方法:想法是取1 , K – 1次,取N – K + 1次。这些数字的总和等于N,并且这些数字的平方和总是最大。对于任意两个非负数a和b , (a 2 + b 2 )始终小于1 +(a + b – 1) 2 。
下面是上述方法的实现:
C++
// C++ program to find K numbers
// with sum equal to N and the
// sum of their squares maximized
#include
using namespace std;
// Function that prints the
// required K numbers
void printKNumbers(int N, int K)
{
// Print 1, K-1 times
for (int i = 0; i < K - 1; i++)
cout << 1 << " ";
// Print (N-K+1)
cout << (N - K + 1);
}
// Driver Code
int main()
{
int N = 10, K = 3;
printKNumbers(N, K);
return 0;
}
Java
// Java program to find K numbers
// with sum equal to N and the
// sum of their squares maximized
class GFG{
// Function that prints the
// required K numbers
static void printKNumbers(int N, int K)
{
// Print 1, K-1 times
for(int i = 0; i < K - 1; i++)
System.out.print(1 + " ");
// Print (N - K + 1)
System.out.print(N - K + 1);
}
// Driver Code
public static void main(String[] args)
{
int N = 10, K = 3;
printKNumbers(N, K);
}
}
// This code is contributed by Amit Katiyar
Python3
# Python3 program to find K numbers
# with a sum equal to N and the
# sum of their squares maximized
# Function that prints the
# required K numbers
def printKNumbers(N, K):
# Print 1, K-1 times
for i in range(K - 1):
print(1, end = ' ')
# Print (N-K+1)
print(N - K + 1)
# Driver code
if __name__=='__main__':
(N, K) = (10, 3)
printKNumbers(N, K)
# This code is contributed by rutvik_56
C#
// C# program to find K numbers
// with sum equal to N and the
// sum of their squares maximized
using System;
class GFG{
// Function that prints the
// required K numbers
static void printKNumbers(int N, int K)
{
// Print 1, K-1 times
for(int i = 0; i < K - 1; i++)
Console.Write(1 + " ");
// Print (N - K + 1)
Console.Write(N - K + 1);
}
// Driver code
public static void Main(String[] args)
{
int N = 10, K = 3;
printKNumbers(N, K);
}
}
// This code is contributed by shivanisinghss2110
Javascript
输出:
1 1 8
时间复杂度: O(K)
辅助空间: O(1)