我们得到了一个数字N。我们的任务是从N生成所有冰雹数字，并找出N采取的步骤数，以减少到
null
科拉茨猜想： L.Collatz在1937年提出的一个问题，也称为3x+1映射，即3n+1问题。设N为整数。根据Collatz猜想，如果我们继续如下迭代N N=N/2//表示偶数N 对于奇数N，N=3*N+1// 不管选择N，我们的数字最终都会收敛到1。 冰雹数量： 科拉兹猜想产生的整数序列称为冰雹数。
例如：
Input : N = 7Output : Hailstone Numbers: 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1No. of steps Required: 17Input : N = 9Output : Hailstone Numbers: 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1No. of steps Required: 20In the first example, N = 7. The numbers will be calculated as follows:73 * 7 + 1 = 22 // Since 7 is odd.22 / 2 = 11 // 22 is even.3 * 11 + 1 = 34 // 11 is odd..... and so on upto 1.
这个想法很简单，我们递归地打印数字，直到达到基本情况。
C++
// C++ program to generate hailstone // numbers and calculate steps required // to reduce them to 1 #include <bits/stdc++.h> using namespace std; // function to print hailstone numbers // and to calculate the number of steps // required int HailstoneNumbers( int N) { static int c; cout << N << " " ; if (N == 1 && c == 0) { // N is initially 1. return c; } else if (N == 1 && c != 0) { // N is reduced to 1. c++; return c; } else if (N % 2 == 0) { // If N is Even. c++; HailstoneNumbers(N / 2); } else if (N % 2 != 0) { // N is Odd. c++; HailstoneNumbers(3 * N + 1); } } // Driver code int main() { int N = 7; int x; // Function to generate Hailstone // Numbers x = HailstoneNumbers(N); // Output: Number of Steps cout << endl; cout << "Number of Steps: " << x; return 0; } |
JAVA
// Java program to generate hailstone // numbers and calculate steps required // to reduce them to 1 import java.util.*; class GFG { static int c; // function to print hailstone numbers // and to calculate the number of steps // required static int HailstoneNumbers( int N) { System.out.print(N + " " ); if (N == 1 && c == 0 ) { // N is initially 1. return c; } else if (N == 1 && c != 0 ) { // N is reduced to 1. c++; return c; } else if (N % 2 == 0 ) { // If N is Even. c++; HailstoneNumbers(N / 2 ); } else if (N % 2 != 0 ) { // N is Odd. c++; HailstoneNumbers( 3 * N + 1 ); } return c; } // Driver code public static void main(String[] args) { int N = 7 ; int x; // Function to generate Hailstone // Numbers x = HailstoneNumbers(N); // Output: Number of Steps System.out.println(); System.out.println( "Number of Steps: " + x); } } /* This code is contributed by Kriti Shukla */ |
python
# Python3 program to generate # hailstone numbers and # calculate steps required # to reduce them to 1 # function to print hailstone # numbers and to calculate # the number of steps required def HailstoneNumbers(N, c): print (N, end = " " ) if (N = = 1 and c = = 0 ): # N is initially 1. return c elif (N = = 1 and c ! = 0 ): # N is reduced to 1. c = c + 1 elif (N % 2 = = 0 ): # If N is Even. c = c + 1 c = HailstoneNumbers( int (N / 2 ), c) elif (N % 2 ! = 0 ): # N is Odd. c = c + 1 c = HailstoneNumbers( 3 * N + 1 , c) return c # Driver Code N = 7 # Function to generate # Hailstone Numbers x = HailstoneNumbers(N, 0 ) # Output: Number of Steps print ( "Number of Steps: " , x) # This code is contributed # by mits |
C#
// C# program to generate hailstone // numbers and calculate steps required // to reduce them to 1 using System; class GFG { static int c; // function to print hailstone numbers // and to calculate the number of steps // required static int HailstoneNumbers( int N) { Console.Write(N + " " ); if (N == 1 && c == 0) { // N is initially 1. return c; } else if (N == 1 && c != 0) { // N is reduced to 1. c++; return c; } else if (N % 2 == 0) { // If N is Even. c++; HailstoneNumbers(N / 2); } else if (N % 2 != 0) { // N is Odd. c++; HailstoneNumbers(3 * N + 1); } return c; } // Driver code public static void Main() { int N = 7; int x; // Function to generate Hailstone // Numbers x = HailstoneNumbers(N); // Output: Number of Steps Console.WriteLine(); Console.WriteLine( "Number of Steps: " + x); } } // This code is contributed by vt_m |
PHP
<?php // PHP program to generate // hailstone numbers and // calculate steps required // to reduce them to 1 // function to print hailstone // numbers and to calculate the // number of steps required function HailstoneNumbers( $N ) { static $c ; echo $N . " " ; if ( $N == 1 && $c == 0) { // N is initially 1. return $c ; } else if ( $N == 1 && $c != 0) { // N is reduced to 1. $c ++; return $c ; } else if ( $N % 2 == 0) { // If N is Even. $c ++; HailstoneNumbers((int)( $N / 2)); } else if ( $N % 2 != 0) { // N is Odd. $c ++; HailstoneNumbers(3 * $N + 1); } return $c ; } // Driver Code $N = 7; // Function to generate // Hailstone Numbers $x = HailstoneNumbers( $N ); // Output: Number of Steps echo "Number of Steps: " . $x ; // This code is contributed // by mits ?> |
Javascript
<script> // JavaScript program to generate hailstone // numbers and calculate steps required // to reduce them to 1 let c = 0; // function to print hailstone numbers // and to calculate the number of steps // required function HailstoneNumbers(N) { document.write(N + " " ); if (N == 1 && c == 0) { // N is initially 1. return c; } else if (N == 1 && c != 0) { // N is reduced to 1. c++; return c; } else if (N % 2 == 0) { // If N is Even. c++; HailstoneNumbers(N / 2); } else if (N % 2 != 0) { // N is Odd. c++; HailstoneNumbers(3 * N + 1); } return c; } // Driver Code let N = 7; let x; // Function to generate Hailstone // Numbers x = HailstoneNumbers(N); // Output: Number of Steps document.write( "<br/>" ); document.write( "Number of Steps: " + x); // This code is contributed by susmitakundugoaldanga. </script> |
输出
7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 Number of Steps: 17
本文由 维尼特·乔希 .如果你喜欢GeekSforgek，并想贡献自己的力量，你也可以使用 写极客。组织 或者把你的文章寄到contribute@geeksforgeeks.org.看到你的文章出现在Geeksforgeks主页上，并帮助其他极客。 如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。
© 版权声明
文章版权归作者所有，未经允许请勿转载。
THE END
![关于”PostgreSQL错误：关系[表]不存在“问题的原因和解决方案-yiteyi-C++库](https://www.yiteyi.com/wp-content/themes/zibll/img/thumbnail.svg)
