找出给定的大正整数乘积中缺失的数字

原文:https://www . geesforgeks . org/find-给定大正整数乘积中缺少的数字/

给定字符串形式的两个大整数 A 和 B 以及字符串形式的它们的乘积 C ，使得乘积的一个数字被 X 替换，任务是在乘积 C 中找到被替换的数字。

示例:

输入: A = 51840，B = 273581，C = 1418243×040 输出: 9 说明: 整数 A 和 B 的乘积为 51840 * 273581 = 14182439040。通过与 C 进行比较，可以得出被替换的数字是 9。因此，打印 9。

输入: A = 123456789，B = 987654321，C = 12193263111×635269 输出: 2

天真方法:解决给定问题的最简单方法是使用本文中讨论的方法找到两个大整数 A 和 B 的乘积，然后将其与 C 进行比较，以找到结果缺失的数字 X 。

时间复杂度:O((log10A)(log10B))* 辅助空间:O(log10A+log10B)

高效方法:上述方法也可以通过使用以下观察值进行优化:

假设 N 是一个整数，N = amam-1am-2……。a2a1a0其中 a x 代表 x 第位数字。现在，N 可以表示为: =>N = am 10m+am-1 10m-1+……+a1* 10+a0

在上式中用 11 执行模运算:

=> N (mod 11) =..+a1(-1)+a0(mod 11)【自 10≦-1(mod 11)】 =>N(mod 11)= T(mod 11)其中 T = a0–a1+a2……+(-1)

因此，从上面的等式中， A * B = C 可以转化为 (A % 11) * (B % 11) = (C % 11) ，其中等式的左手边是一个常数值，右手边将是一个等式中的一个变量 X ，可以求解得到 X 的值。在与 11 进行模运算后， X 可能会有负值，在这种情况下，考虑 X 的正值。

下面是上述方法的实现:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the replaced digit
// in the product of a*b
int findMissingDigit(string a, string b,
                     string c)
{
    // Keeps track of the sign of the
    // current digit
    int w = 1;

    // Stores the value of a % 11
    int a_mod_11 = 0;

    // Find the value of a mod 11 for
    // large value of a as per the
    // derived formula
    for (int i = a.size() - 1; i >= 0; i--) {
        a_mod_11 = (a_mod_11 + w * (a[i] - '0')) % 11;
        w = w * -1;
    }

    // Stores the value of b % 11
    int b_mod_11 = 0;
    w = 1;

    // Find the value of b mod 11 for
    // large value of a as per the
    // derived formula
    for (int i = b.size() - 1;
         i >= 0; i--) {

        b_mod_11 = (b_mod_11
                    + w * (b[i] - '0'))
                   % 11;
        w = w * -1;
    }

    // Stores the value of c % 11
    int c_mod_11 = 0;

    // Keeps track of the sign of x
    bool xSignIsPositive = true;
    w = 1;

    for (int i = c.size() - 1; i >= 0; i--) {

        // If the current digit is the
        // missing digit, then keep
        // the track of its sign
        if (c[i] == 'x') {
            xSignIsPositive = (w == 1);
        }
        else {
            c_mod_11 = (c_mod_11
                        + w * (c[i] - '0'))
                       % 11;
        }
        w = w * -1;
    }

    // Find the value of x using
    // the derived equation
    int x = ((a_mod_11 * b_mod_11)
             - c_mod_11)
            % 11;

    // Check if x has a negative sign
    if (!xSignIsPositive) {
        x = -x;
    }

    // Return positive equivaluent
    // of x mod 11
    return (x % 11 + 11) % 11;
}

// Driver Code
int main()
{
    string A = "123456789";
    string B = "987654321";
    string C = "12193263111x635269";
    cout << findMissingDigit(A, B, C);

    return 0;
}

Java 语言(一种计算机语言，尤用于创建网站)

// Java program for the above approach
class GFG {

    // Function to find the replaced digit
    // in the product of a*b
    public static int findMissingDigit(String a, String b, String c)
    {

        // Keeps track of the sign of the
        // current digit
        int w = 1;

        // Stores the value of a % 11
        int a_mod_11 = 0;

        // Find the value of a mod 11 for
        // large value of a as per the
        // derived formula
        for (int i = a.length() - 1; i >= 0; i--) {
            a_mod_11 = (a_mod_11 + w * (a.charAt(i) - '0')) % 11;
            w = w * -1;
        }

        // Stores the value of b % 11
        int b_mod_11 = 0;
        w = 1;

        // Find the value of b mod 11 for
        // large value of a as per the
        // derived formula
        for (int i = b.length() - 1; i >= 0; i--) {

            b_mod_11 = (b_mod_11 + w * (b.charAt(i) - '0')) % 11;
            w = w * -1;
        }

        // Stores the value of c % 11
        int c_mod_11 = 0;

        // Keeps track of the sign of x
        boolean xSignIsPositive = true;
        w = 1;

        for (int i = c.length() - 1; i >= 0; i--) {

            // If the current digit is the
            // missing digit, then keep
            // the track of its sign
            if (c.charAt(i) == 'x') {
                xSignIsPositive = (w == 1);
            } else {
                c_mod_11 = (c_mod_11 + w * (c.charAt(i) - '0')) % 11;
            }
            w = w * -1;
        }

        // Find the value of x using
        // the derived equation
        int x = ((a_mod_11 * b_mod_11) - c_mod_11) % 11;

        // Check if x has a negative sign
        if (!xSignIsPositive) {
            x = -x;
        }

        // Return positive equivaluent
        // of x mod 11
        return (x % 11 + 11) % 11;
    }

    // Driver Code
    public static void main(String args[]) {
        String A = "123456789";
        String B = "987654321";
        String C = "12193263111x635269";
        System.out.println(findMissingDigit(A, B, C));

    }
}

// This code is contributed by saurabh_jaiswal.

Python 3

# Python3 Program to implement the above approach

# Function to find the replaced digit
# in the product of a*b
def findMissingDigit(a, b, c):

    # Keeps track of the sign of the
    # current digit
    w = 1

    # Stores the value of a % 11
    a_mod_11 = 0

    # Find the value of a mod 11 for
    # large value of a as per the
    # derived formula
    for i in range(len(a) - 1, -1, -1):
        a_mod_11 = (a_mod_11 + w * (ord(a[i]) - ord('0'))) % 11
        w = w * -1

    # Stores the value of b % 11
    b_mod_11 = 0
    w = 1

    # Find the value of b mod 11 for
    # large value of a as per the
    # derived formula
    for i in range(len(b) - 1, -1, -1):
        b_mod_11 = (b_mod_11 + w * (ord(b[i]) - ord('0'))) % 11
        w = w * -1

    # Stores the value of c % 11
    c_mod_11 = 0

    # Keeps track of the sign of x
    xSignIsPositive = True
    w = 1

    for i in range(len(c) - 1, -1, -1):
        # If the current digit is the
        # missing digit, then keep
        # the track of its sign
        if (c[i] == 'x'):
            xSignIsPositive = (w == 1)
        else:
            c_mod_11 = (c_mod_11 + w * (ord(c[i]) - ord('0'))) % 11
        w = w * -1

    # Find the value of x using
    # the derived equation
    x = ((a_mod_11 * b_mod_11) - c_mod_11) % 11

    # Check if x has a negative sign
    if (not xSignIsPositive):
        x = -x

    # Return positive equivaluent
    # of x mod 11
    return (x % 11 + 11) % 11

A = "123456789"
B = "987654321"
C = "12193263111x635269"
print(findMissingDigit(A, B, C))

# This code is contributed by divyeshrabadiya07.

C

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to find the replaced digit
// in the product of a*b
static int findMissingDigit(string a, string b,
                     string c)
{

    // Keeps track of the sign of the
    // current digit
    int w = 1;

    // Stores the value of a % 11
    int a_mod_11 = 0;

    // Find the value of a mod 11 for
    // large value of a as per the
    // derived formula
    for (int i = a.Length - 1; i >= 0; i--) {
        a_mod_11 = (a_mod_11 + w * ((int)a[i] - 48)) % 11;
        w = w * -1;
    }

    // Stores the value of b % 11
    int b_mod_11 = 0;
    w = 1;

    // Find the value of b mod 11 for
    // large value of a as per the
    // derived formula
    for (int i = b.Length - 1;
         i >= 0; i--) {

        b_mod_11 = (b_mod_11
                    + w * ((int)b[i] - 48))
                   % 11;
        w = w * -1;
    }

    // Stores the value of c % 11
    int c_mod_11 = 0;

    // Keeps track of the sign of x
    bool xSignIsPositive = true;
    w = 1;

    for (int i = c.Length - 1; i >= 0; i--) {

        // If the current digit is the
        // missing digit, then keep
        // the track of its sign
        if (c[i] == 'x') {
            xSignIsPositive = (w == 1);
        }
        else {
            c_mod_11 = (c_mod_11
                        + w * ((int)c[i] - '0'))
                       % 11;
        }
        w = w * -1;
    }

    // Find the value of x using
    // the derived equation
    int x = ((a_mod_11 * b_mod_11)
             - c_mod_11)
            % 11;

    // Check if x has a negative sign
    if (xSignIsPositive == false) {
        x = -x;
    }

    // Return positive equivaluent
    // of x mod 11
    return (x % 11 + 11) % 11;
}

// Driver Code
public static void Main()
{
    string A = "123456789";
    string B = "987654321";
    string C = "12193263111x635269";
    Console.Write(findMissingDigit(A, B, C));
}
}

// This code is contributed by ipg2016107.

java 描述语言

<script>
        // JavaScript Program to implement
        // the above approach

        // Function to find the replaced digit
        // in the product of a*b
        function findMissingDigit(a, b,
            c)
        {
            // Keeps track of the sign of the
            // current digit
            let w = 1;

            // Stores the value of a % 11
            let a_mod_11 = 0;

            // Find the value of a mod 11 for
            // large value of a as per the
            // derived formula
            for (let i = a.length - 1; i >= 0; i--) {
                a_mod_11 = (a_mod_11 + w * (a[i].charCodeAt(0) - '0'.charCodeAt(0))) % 11;
                w = w * -1;
            }

            // Stores the value of b % 11
            let b_mod_11 = 0;
            w = 1;

            // Find the value of b mod 11 for
            // large value of a as per the
            // derived formula
            for (let i = b.length - 1;
                i >= 0; i--) {

                b_mod_11 = (b_mod_11
                    + w * (b[i].charCodeAt(0) - '0'.charCodeAt(0)))
                    % 11;
                w = w * -1;
            }

            // Stores the value of c % 11
            let c_mod_11 = 0;

            // Keeps track of the sign of x
            let xSignIsPositive = true;
            w = 1;

            for (let i = c.length - 1; i >= 0; i--) {

                // If the current digit is the
                // missing digit, then keep
                // the track of its sign
                if (c[i] == 'x') {
                    xSignIsPositive = (w == 1);
                }
                else {
                    c_mod_11 = (c_mod_11
                        + w * (c[i].charCodeAt(0) - '0'.charCodeAt(0)))
                        % 11;
                }
                w = w * -1;
            }

            // Find the value of x using
            // the derived equation
            let x = ((a_mod_11 * b_mod_11)
                - c_mod_11)
                % 11;

            // Check if x has a negative sign
            if (!xSignIsPositive) {
                x = -x;
            }

            // Return positive equivaluent
            // of x mod 11
            return (x % 11 + 11) % 11;
        }

        // Driver Code
        let A = "123456789";
        let B = "987654321";
        let C = "12193263111x635269";
        document.write(findMissingDigit(A, B, C));

// This code is contributed by Potta Lokesh
</script>

Output: 

2

时间复杂度:O(log10A+log10B) 辅助空间: O(1)

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