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Package com; public class test {public static void main (string [] ARGs) {system. out. println (getsteps1 (); system. out. println (">>>>>>>>>>>>>>>>>>>>>>>>>>>>"); system. out. println (test. getsteps (); system. out. println (">>>>>>>>>>>>>>>>>>>>>>>>>>>> ");} public static int getsteps () {// use the minimum public multiple to reduce the number of traversal times. Int I = 1; int step = 2; Boolean maxstep

#include Be careful to define its type, divide it into two parts, and define it as "I" to see if the denominator is an odd or even number, and the sum is summed. C language: Calculate 1/1-1/2+1/3-1/4+

There is a fractional sequence of 2/1 + 3/2 + 5/3 + 8/5 + 13/8 +... Find the sum of the first 20 items of the series.Program: # Include Output result: 32.660261 Press any key to continue

C language: There is a score sequence: 2/1 + 3/2 + 5/3 + 8/5 + 13/8 +... Find the sum of the first 20 items in this seriesProgram: # Include Output result: 32.660261 Press any key to continue

In order to omit the space and make it visible to the operator who manually fills in the paper, the volume number on the card of the materials shelf is determined to be classified and sorted, as shown in A-3, A-4, A-5, A-8 forming A-3 ~ 5, 8, etc. The following code uses a few auxiliary list /// /// Similar to 1, 2, 3

View code //// Main. M // money /// enter a specified amount (in Yuan, for example, 345.78) from the keyboard, and then display the number of different denominations that pay the amount, required to display 100 yuan, 50 yuan, 10 yuan, 5 yuan, 2 yuan, 1 yuan, 5 cents, 1 cent

/**5-1* Define interface printable, which includes a method Printitmyway (),* This method has no formal parameters and the return value is null**/Interface Printable{void Printitmyway ();}/**5-2* Rewrite the rectangle class in experiment 3 to implement the printable interface,* Use the Printitmyway () method to relate

/***//** * Fractionserial. Java * There is a fractional sequence: 2/1, 3/2, 5/3, 8/5, 13/8, 21/13... * Calculate the sum of the first 20 items of the series. * @ Author Deng Chao (codingmouse) * @ Version 0.2 * Development/test environment: jdk1.6 + eclipse SDK 3.3.2 */ Pub

Calculate the sum of the First n items of the Fibonacci fractional sequence (N is a constant, and the Fibonacci fractional sequence is 2/1, 3/2, 5/3, 8/5 ,...). # Include Stdio. h > # Include Conio. h > Void Main (){ Int I, N; Float F1 =

# Include }/* The numerator behind the score is equal to the numerator plus the denominator of the previous score, and the denominator of the subsequent score is equal to the numerator with the previous score */ There is a fractional sequence: 2/1, 3/2, 5/3, 8/5, 13/8, 21

Package COM. WZS; // Add difficulty to the first question. Use the numbers 1, 2, 3, 4, and 5 to write a main function in Java and print out all the different orders, // For example, 51234, 12345, etc., the requirement: "4" cannot be In the third place, "3" and "5" cannot be connected. Public class test3 {public static