dsal 求平方加 X 时不以 T1 或 T2 结尾的整数

求平方加 X 时不以 T1 或 T2 结尾的整数

原文:https://www . geesforgeks . org/find-the-integers-not-end-with-T1-or-T2-when-squared-and-add-x/

给定一组 N 个整数。给定两个一位数的数字 T1 和 T2 以及一个数字 X。任务是找出其中不以 T1 或 T2 结束的整数，当它们被平方并且 X 被加到它们上面时。如果没有这样的整数，打印 -1 。 举例:

输入: N = 4，arr[] = {3，1，4，7} X = 10，T1 = 5，T2 = 6 输出: 19 11 59 解释: 的修改值 3 为 19 (3^2 + 10)。 1的修正值为 11 (1^2 + 10)。 4的修正值为 26 (4^2 + 10)。 7 的修正值为 59 (7^2 + 10)。不以 5 或 6 结尾的修改值为 19、11、59 。因此输出为 19 11 59 。 输入: N = 4，arr[] = {2，18，22，8} X = 2，T1 = 5，T2 = 6 输出: -1 解释: 的修改值 2 为 6 (2^2 + 2)。 18 的修正值为 326 (18^2 + 2)。 22的修正值为 486 (22^2 + 2)。 8的修正值为 66 (8^2 + 2)。 As，没有以 5 或 6 结尾的修改值。因此输出为 -1 。

进场:

初始化一个布尔变量标志为真。
遍历数组中的元素a【n】。
将 X 和【I】的平方之和存储在变量 temp 中。
检查温度的最后一位数字是不是既不是 T1 也不是 T2 。
如果是，则打印温度中的值，并将标志更改为假。
遍历完数组中的所有元素后，如果标志为真，则打印 -1 。

以下是上述方法的实现:

C++

// C++ program to find the integers
// that ends with either T1 or T2
// when squared and added X

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

// Function to print the elements
// Not ending with T1 or T2
void findIntegers(int n, int a[],
                   int x, int t1, int t2)
{

    // Flag to check if none of the elements
    // Do not end with t1 or t2
    bool flag = true;

    // Traverse through all the elements
    for (int i = 0; i < n; i++) {

        // Temporary variable to store the value
        int temp = pow(a[i], 2) + x;

        // If the last digit is neither t1
        // nor t2 then
        if (temp % 10 != t1 && temp % 10 != t2) {

            // Print the number
            cout << temp << " ";

            // Set the flag as False
            flag = false;
        }
    }

    // If none of the elements
    // meets the specification
    if (flag)
        cout << "-1";
}

// Driver Code
int main()
{
    // Test case 1
    int N = 4, X = 10, T1 = 5, T2 = 6;
    int a[N] = { 3, 1, 4, 7 };

    // Call the function
    findIntegers(N, a, X, T1, T2);
    cout << endl;

    // Test case 2
    N = 4, X = 2, T1 = 5, T2 = 6;
    int b[N] = { 2, 18, 22, 8 };

    // Call the function
    findIntegers(N, b, X, T1, T2);

    return 0;
}

Java 语言(一种计算机语言，尤用于创建网站)

// Java program to find the integers
// that ends with either T1 or T2
// when squared and added X
class GFG
{

    // Function to print the elements
    // Not ending with T1 or T2
    static void findIntegers(int n, int a[],
                             int x, int t1, int t2)
    {

        // Flag to check if none of the elements
        // Do not end with t1 or t2
        boolean flag = true;

        // Traverse through all the elements
        for (int i = 0; i < n; i++)
        {

            // Temporary variable to store the value
            int temp = (int)Math.pow(a[i], 2) + x;

            // If the last digit is neither t1
            // nor t2 then
            if (temp % 10 != t1 && temp % 10 != t2)
            {

                // Print the number
                System.out.print(temp + " ");

                // Set the flag as False
                flag = false;
            }
        }

        // If none of the elements
        // meets the specification
        if (flag)
        {
            System.out.println();
            System.out.print("-1");
        }
    }

    // Driver Code
    public static void main(String args[])
    {
        // Test case 1
        int N = 4;
        int X = 10;
        int T1 = 5;
        int T2 = 6;
        int a[] = { 3, 1, 4, 7 };

        // Call the function
        findIntegers(N, a, X, T1, T2);

        // Test case 2
        N = 4; X = 2; T1 = 5; T2 = 6;
        int b[] = { 2, 18, 22, 8 };

        // Call the function
        findIntegers(N, b, X, T1, T2);
    }
}

// This code is contributed by AnkitRai01

Python 3

# Python3 program to find the integers
# that ends with either T1 or T2
# when squared and added X

# Function to print the elements
# Not ending with T1 or T2
def findIntegers(n, a, x, t1, t2):

    # Flag to check if none of the elements
    # Do not end with t1 or t2
    flag = True

    # Traverse through all the elements
    for i in range(n):

        # Temporary variable to store the value
        temp = pow(a[i], 2) + x

        # If the last digit is neither t1
        # nor t2 then
        if(temp % 10 != t1 and
           temp % 10 != t2):

            # Print the number
            print(temp, end = " ")

            # Set the flag as False
            flag = False

    # If none of the elements
    # meets the specification
    if flag:
        print(-1)

# Driver Code

# Test case 1
N , X , T1 , T2 = 4 , 10 , 5 , 6
a = [ 3, 1, 4, 7 ]

# Call the function
findIntegers(N, a, X, T1, T2);
print()

# Test case 2
N , X , T1 , T2 = 4 , 2 , 5 , 6
b = [ 2, 18, 22, 8 ]

# Call the function
findIntegers(N, b, X, T1, T2)

# This code is contributed by divyamohan123

C

// C# program to find the integers
// that ends with either T1 or T2
// when squared and added X
using System;

class GFG
{

    // Function to print the elements
    // Not ending with T1 or T2
    static void findIntegers(int n, int []a,
                             int x, int t1, int t2)
    {

        // Flag to check if none of the elements
        // Do not end with t1 or t2
        bool flag = true;

        // Traverse through all the elements
        for (int i = 0; i < n; i++)
        {

            // Temporary variable to store the value
            int temp = (int)Math.Pow(a[i], 2) + x;

            // If the last digit is neither t1
            // nor t2 then
            if (temp % 10 != t1 &&
                temp % 10 != t2)
            {

                // Print the number
                Console.Write(temp + " ");

                // Set the flag as False
                flag = false;
            }
        }

        // If none of the elements
        // meets the specification
        if (flag)
        {
            Console.WriteLine();
            Console.Write("-1");
        }
    }

    // Driver Code
    public static void Main(String []args)
    {
        // Test case 1
        int N = 4;
        int X = 10;
        int T1 = 5;
        int T2 = 6;
        int []a = { 3, 1, 4, 7 };

        // Call the function
        findIntegers(N, a, X, T1, T2);

        // Test case 2
        N = 4; X = 2; T1 = 5; T2 = 6;
        int []b = { 2, 18, 22, 8 };

        // Call the function
        findIntegers(N, b, X, T1, T2);
    }
}

// This code is contributed by 29AjayKumar

java 描述语言

<script>
// Javascript program to find the integers
// that ends with either T1 or T2
// when squared and added X

// Function to print the elements
// Not ending with T1 or T2
function findIntegers(n, a, x, t1, t2)
{

    // Flag to check if none of the elements
    // Do not end with t1 or t2
    let flag = true;

    // Traverse through all the elements
    for (let i = 0; i < n; i++) {

        // Temporary variable to store the value
        let temp = Math.pow(a[i], 2) + x;

        // If the last digit is neither t1
        // nor t2 then
        if (temp % 10 != t1 && temp % 10 != t2) {

            // Print the number
            document.write(temp + " ");

            // Set the flag as False
            flag = false;
        }
    }

    // If none of the elements
    // meets the specification
    if (flag)
        document.write("-1");
}

// Driver Code
    // Test case 1
    let N = 4, X = 10, T1 = 5, T2 = 6;
    let a = [ 3, 1, 4, 7 ];

    // Call the function
    findIntegers(N, a, X, T1, T2);
    document.write("<br>");

    // Test case 2
    N = 4, X = 2, T1 = 5, T2 = 6;
    let b = [ 2, 18, 22, 8 ];

    // Call the function
    findIntegers(N, b, X, T1, T2);

</script>

Output:

19 11 59 
-1

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dsal 求平方加 X 时不以 T1 或 T2 结尾的整数

求平方加 X 时不以 T1 或 T2 结尾的整数

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《GeeksForGeeks 数据结构和算法中文教程 2022-10-24》

dsal 求平方加 X 时不以 T1 或 T2 结尾的整数

求平方加 X 时不以 T1 或 T2 结尾的整数

C++

Java 语言(一种计算机语言，尤用于创建网站)

Python 3

C

java 描述语言

留言

联系客服

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