给定两个由N个整数组成的数组。考虑一个数组C ，其中第i个整数将是d * a [i] + b [i] ，其中d是任何任意实数。任务是打印d，使数组C的最大零个数也打印零个。
例子：
Input: a[] = {1, 2, 3, 4, 5}, b[] = {2, 4, 7, 11, 3}
Output:
Value of d is: -2
The number of zeros in array C is: 2
If we choose d as -2 then we get two zeros in the array C which is the maximum possible.
Input: a[] = {13, 37, 39} b[] = {1, 2, 3}
Output:
Value of d is: -0.0769231
The number of zeros in array C is:
可以按照以下步骤解决上述问题：
- 该方程可重写为d = -b [i] / a [i]
- 使用哈希表计算任何实数的最大出现次数，以得到d的值。
- 零数将是最大计数+(对数a [i]和b [i]均为零)。
下面是上述方法的实现：
C++
// C++ program to implement the above
// approach
#include
using namespace std;
// Function to find the value of d
// and find the number of zeros in the array
void findDandZeros(int a[], int b[], int n)
{
// Hash table
unordered_map mpp;
int count = 0;
// Iterate for i-th element
for (int i = 0; i < n; i++) {
// If both are not 0
if (b[i] != 0 && a[i] != 0) {
long double val = (long double)(-1.0 * b[i]) /
(long double)(a[i]);
mpp[val] += 1;
}
// If both are 0
else if (b[i] == 0 && a[i] == 0)
count += 1;
}
// Find max occurring d
int maxi = 0;
for (auto it : mpp) {
maxi = max(it.second, maxi);
}
// Print the d which occurs max times
for (auto it : mpp) {
if (it.second == maxi) {
cout << "Value of d is: "
<< it.first << endl;
break;
}
}
// Print the number of zeros
cout << "The number of zeros in array C is: "
<< maxi + count;
}
// Driver code
int main()
{
int a[] = { 13, 37, 39 };
int b[] = { 1, 2, 3 };
int n = sizeof(a) / sizeof(a[0]);
findDandZeros(a, b, n);
return 0;
} Java
// Java program to implement the above
// approach
import java.util.*;
class geeks
{
// Function to find the value of d
// and find the number of zeros in the array
public static void findDandZeroes(int[] a, int[] b, int n)
{
// Hash table
HashMap mpp = new HashMap<>();
int count = 0;
// Iterate for i-th element
for (int i = 0; i < n; i++)
{
// If both are not 0
if (b[i] != 0 && a[i] != 0)
{
double val = (double) (-1.0 * b[i]) / (double) (a[i]);
if (mpp.get(val) != null)
{
int x = mpp.get(val);
mpp.put(val, ++x);
}
else
mpp.put(val, 1);
}
// If both are 0
else if (b[i] == 0 && a[i] == 0)
count += 1;
}
// Find max occurring d
int maxi = 0;
for (HashMap.Entry entry : mpp.entrySet())
{
maxi = Math.max(entry.getValue(), maxi);
}
// Print the d which occurs max times
for (HashMap.Entry entry : mpp.entrySet())
{
if (entry.getValue() == maxi)
{
System.out.println("Value of d is: " + entry.getKey());
break;
}
}
// Print the number of zeros
System.out.println("The number of zeros in array C is: " +
(maxi + count));
}
// Driver Code
public static void main(String[] args)
{
int[] a = { 13, 37, 39 };
int[] b = { 1, 2, 3 };
int n = a.length;
findDandZeroes(a, b, n);
}
}
// This code is contributed by
// sanjeev2552 Python3
# Python3 program to implement the
# above approach
# Function to find the value of d and
# find the number of zeros in the array
def findDandZeros(a, b, n) :
# Hash table
mpp = {};
count = 0;
# Iterate for i-th element
for i in range(n) :
# If both are not 0
if (b[i] != 0 and a[i] != 0) :
val = (-1.0 * b[i]) / a[i];
if val not in mpp :
mpp[val] = 0;
mpp[val] += 1;
# If both are 0
elif (b[i] == 0 and a[i] == 0) :
count += 1;
# Find max occurring d
maxi = 0;
for item in mpp :
maxi = max(mpp[item], maxi);
# Print the d which occurs max times
for keys, values in mpp.items() :
if (values == maxi) :
print("Value of d is:", keys);
break;
# Print the number of zeros
print("The number of zeros in array C is:",
maxi + count);
# Driver code
if __name__ == "__main__" :
a = [ 13, 37, 39 ];
b = [ 1, 2, 3 ];
n = len(a);
findDandZeros(a, b, n);
# This code is contributed by RyugaC#
// C# program to implement the above
// approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the value of d
// and find the number of zeros in the array
public static void findDandZeroes(int[] a,
int[] b, int n)
{
// Hash table
Dictionary mpp = new Dictionary();
int count = 0;
// Iterate for i-th element
for (int i = 0; i < n; i++)
{
// If both are not 0
if (b[i] != 0 && a[i] != 0)
{
double val = (double)(-1.0 * b[i]) /
(double)(a[i]);
if (mpp.ContainsKey(val))
{
mpp[val] = ++mpp[val];
}
else
mpp.Add(val, 1);
}
// If both are 0
else if (b[i] == 0 && a[i] == 0)
count += 1;
}
// Find max occurring d
int maxi = 0;
foreach(KeyValuePair entry in mpp)
{
maxi = Math.Max(entry.Value, maxi);
}
// Print the d which occurs max times
foreach(KeyValuePair entry in mpp)
{
if (entry.Value == maxi)
{
Console.WriteLine("Value of d is: " +
entry.Key);
break;
}
}
// Print the number of zeros
Console.WriteLine("The number of zeros in array C is: " +
(maxi + count));
}
// Driver Code
public static void Main(String[] args)
{
int[] a = { 13, 37, 39 };
int[] b = { 1, 2, 3 };
int n = a.Length;
findDandZeroes(a, b, n);
}
}
// This code is contributed by Rajput-Ji 输出：
Value of d is: -0.0769231
The number of zeros in array C is: 2
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