求数组乘法除以 n 的余数

给定多个数字和一个数字 n，任务是打印所有数字除以 n 后的余数。
例子：

Input : arr[] = {100, 10, 5, 25, 35, 14}, 
            n = 11
Output : 9
100 x 10 x 5 x 25 x 35 x 14 = 61250000 % 11 = 9

Input : arr[] = {100, 10}, 
            n = 5 
Output : 0
100 x 10 = 1000 % 5 = 0

天真的方法：首先将所有数字相乘，然后将 % 乘以 n，然后找到余数，但在这种方法中，如果数字最大为 2^64，那么它会给出错误的答案。
避免溢出的方法：首先取一个余数或单个数字，如 arr[i] % n。然后将余数乘以当前结果。乘法后，再次取余数以避免溢出。这是因为模算术的分布特性。 ( a * b) % c = ( ( a % c ) * ( b % c ) ) % c

C++

// C++ program to find
// remainder when all
// array elements are
// multiplied.
#include 
using namespace std;
 
// Find remainder of arr[0] * arr[1] *
// .. * arr[n-1]
int findremainder(int arr[], int len, int n)
{
    int mul = 1;
 
    // find the individual remainder
    // and multiple with mul.
    for (int i = 0; i < len; i++)
        mul = (mul * (arr[i] % n)) % n;
     
    return mul % n;
}
 
// Driver code
int main()
{
    int arr[] = { 100, 10, 5, 25, 35, 14 };
    int len = sizeof(arr) / sizeof(arr[0]);
    int n = 11;
 
    // print the remainder of after
    // multiple all the numbers
    cout << findremainder(arr, len, n);
}

Java

// Java program to find
// remainder when all
// array elements are
// multiplied.
import java.util.*;
import java.lang.*;
 
public class GfG{
     
    // Find remainder of arr[0] * arr[1] *
    // .. * arr[n-1]
    public static int findremainder(int arr[],
                                   int len, int n)
    {
        int mul = 1;
 
        // find the individual remainder
        // and multiple with mul.
        for (int i = 0; i < len; i++)
            mul = (mul * (arr[i] % n)) % n;
     
        return mul % n;
    }
     
    // Driver function
    public static void main(String argc[])
    {
        int[] arr = new int []{ 100, 10, 5,
                                25, 35, 14 };
        int len = 6;
        int n = 11;
 
        // print the remainder of after
        // multiple all the numbers
        System.out.println(findremainder(arr, len, n));
    }
}
 
/* This code is contributed by Sagar Shukla */

Python3

# Python3 program to
# find remainder when
# all array elements
# are multiplied.
 
# Find remainder of arr[0] * arr[1]
# * .. * arr[n-1]
def findremainder(arr, lens, n):
    mul = 1
 
    # find the individual
    # remainder and
    # multiple with mul.
    for i in range(lens):
        mul = (mul * (arr[i] % n)) % n
     
    return mul % n
 
# Driven code
arr = [ 100, 10, 5, 25, 35, 14 ]
lens = len(arr)
n = 11
 
# print the remainder
# of after multiple
# all the numbers
print( findremainder(arr, lens, n))
 
# This code is contributed by "rishabh_jain".

C#

// C# program to find
// remainder when all
// array elements are
// multiplied.
using System;
 
public class GfG{
     
    // Find remainder of arr[0] * arr[1] *
    // .. * arr[n-1]
    public static int findremainder(int []arr,
                                int len, int n)
    {
        int mul = 1;
 
        // find the individual remainder
        // and multiple with mul.
        for (int i = 0; i < len; i++)
            mul = (mul * (arr[i] % n)) % n;
     
        return mul % n;
    }
     
    // Driver function
    public static void Main()
    {
        int[] arr = new int []{ 100, 10, 5,
                                25, 35, 14 };
        int len = 6;
        int n = 11;
 
        // print the remainder of after
        // multiple all the numbers
        Console.WriteLine(findremainder(arr, len, n));
    }
}
 
/* This code is contributed by vt_m */

Javascript

    // Javascript program to find
    // remainder when all
    // array elements are
    // multiplied.
     
    // Find remainder of arr[0] * arr[1] *
    // .. * arr[n-1]
    function findremainder(arr, len, n)
    {
        let mul = 1;
   
        // find the individual remainder
        // and multiple with mul.
        for (let i = 0; i < len; i++)
            mul = (mul * (arr[i] % n)) % n;
       
        return mul % n;
    }
     
    let arr = [ 100, 10, 5, 25, 35, 14 ];
    let len = 6;
    let n = 11;
 
    // print the remainder of after
    // multiple all the numbers
    document.write(findremainder(arr, len, n));
 
// This code is contributed by rameshtravel07.

输出：

9

时间复杂度： O(len)，其中 len 是给定数组的大小
辅助空间： O(1)