求两个数乘积的最大位数之和
给定一个大小为 N ( > 2)的数组 arr[] 。任务是找出给定数组中任意两个数乘积的最大位数和。 例:
输入: arr[] = {8，7} 输出:11 8与 7 的乘积为 56 。56 位数之和等于 11 。 输入: arr[] = {4，3，5} 输出:6 4&3 的乘积= 12。数字之和= 3。 3 的乘积& 5 = 15。数字总和= 6。 4 的乘积& 5 = 20。数字之和= 2。
方法:运行嵌套循环，选择数组的两个数，得到乘积。对于每个产品，检查数字总和并找出最大数字总和。 以下是上述方法的实施:
C++
// C++ program find the maximum sum of
// digits of the product of two numbers
#include <bits/stdc++.h>
using namespace std;
// Function to find the sum of the digits
int sumDigits(int n)
{
int digit_sum = 0;
while (n) {
digit_sum += n % 10;
n /= 10;
}
return digit_sum;
}
// Function to find the maximum sum of digits of product
int productOfNumbers(int arr[], int n)
{
int sum = INT_MIN;
// Run nested loops
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
int product = arr[i] * arr[j];
// Find the maximum sum
sum = max(sum, sumDigits(product));
}
}
// Return the required answer
return sum;
}
// Driver code
int main()
{
int arr[] = { 4, 3, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << productOfNumbers(arr, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program find the maximum sum of
// digits of the product of two numbers
import java.io.*;
class GFG
{
// Function to find the sum of the digits
static int sumDigits(int n)
{
int digit_sum = 0;
while (n > 0)
{
digit_sum += n % 10;
n /= 10;
}
return digit_sum;
}
// Function to find the maximum sum
// of digits of product
static int productOfNumbers(int []arr, int n)
{
int sum = Integer.MIN_VALUE;
// Run nested loops
for (int i = 0; i < n - 1; i++)
{
for (int j = i + 1; j < n; j++)
{
int product = arr[i] * arr[j];
// Find the maximum sum
sum = Math.max(sum, sumDigits(product));
}
}
// Return the required answer
return sum;
}
// Driver code
public static void main (String[] args)
{
int []arr = { 4, 3, 5 };
int n = arr.length;
System.out.print( productOfNumbers(arr, n));
}
}
// This code is contributed by anuj_67..
Python 3
# Python3 program find the maximum sum of
# digits of the product of two numbers
import sys
# Function to find the sum of the digits
def sumDigits(n):
digit_sum = 0;
while (n > 0):
digit_sum += n % 10;
n /= 10;
return digit_sum;
# Function to find the maximum sum
# of digits of product
def productOfNumbers(arr, n):
sum = -sys.maxsize - 1;
# Run nested loops
for i in range(n - 1):
for j in range(i + 1, n):
product = arr[i] * arr[j];
# Find the maximum sum
sum = max(sum, sumDigits(product));
# Return the required answer
return sum;
# Driver code
if __name__ == '__main__':
arr =[ 4, 3, 5 ];
n = len(arr);
print(int(productOfNumbers(arr, n)));
# This code contributed by PrinciRaj1992
C
// C# program find the maximum sum of
// digits of the product of two numbers
using System;
class GFG
{
// Function to find the sum of the digits
static int sumDigits(int n)
{
int digit_sum = 0;
while (n > 0)
{
digit_sum += n % 10;
n /= 10;
}
return digit_sum;
}
// Function to find the maximum sum
// of digits of product
static int productOfNumbers(int []arr, int n)
{
int sum = int.MinValue;
// Run nested loops
for (int i = 0; i < n - 1; i++)
{
for (int j = i + 1; j < n; j++)
{
int product = arr[i] * arr[j];
// Find the maximum sum
sum = Math.Max(sum, sumDigits(product));
}
}
// Return the required answer
return sum;
}
// Driver code
public static void Main (String[] args)
{
int []arr = { 4, 3, 5 };
int n = arr.Length;
Console.Write(productOfNumbers(arr, n));
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript program find the maximum sum of
// digits of the product of two numbers
// Function to find the sum of the digits
function sumDigits(n)
{
let digit_sum = 0;
while (n > 0) {
digit_sum += n % 10;
n = parseInt(n / 10, 10);
}
return digit_sum;
}
// Function to find the maximum sum of digits of product
function productOfNumbers(arr, n)
{
let sum = Number.MIN_VALUE;
// Run nested loops
for (let i = 0; i < n - 1; i++) {
for (let j = i + 1; j < n; j++) {
let product = arr[i] * arr[j];
// Find the maximum sum
sum = Math.max(sum, sumDigits(product));
}
}
// Return the required answer
return sum;
}
let arr = [ 4, 3, 5 ];
let n = arr.length;
document.write(productOfNumbers(arr, n));
</script>
输出:
6
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