Given a number N, the task is to find the N integer values of Xi such that X1 < X2 < … < XN and sin(X1) < sin(X2) < … < sin(XN).
Examples:
Input: N = 5
Output:
X1 = 0 sin(X1) = 0.000000
X2 = 710 sin(X2) = 0.000060
X3 = 1420 sin(X3) = 0.000121
X4 = 2130 sin(X4) = 0.000181
X5 = 2840 sin(X5) = 0.000241
Input: N = 3
Output:
X1 = 0 sin(X1) = 0.000000
X2 = 710 sin(X2) = 0.000060
X3 = 1420 sin(X3) = 0.000121
Approach: The idea is to use the fractional value of PI(&PI); i.e., Pi = 355/113 as it given best rational value of PI of accuracy being 0.000009%.
As,
PI = 355/113
=> 113*PI = 355
=> 2*(113*PI) = 710
As sin() function has a period of 2*PI,
Therefore sin(2*k*PI + Y) = sin(Y);
As per the above equation to get the value X1 < X2 < … < XN and sin(X1) < sin(X2) < … < sin(XN) we must find the value of sin(X) with an increment of 710.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to print all such Xi s.t.// all Xi and sin(Xi) are strictly// increasingvoid printSinX(int N){ int Xi = 0; int num = 1; // Till N becomes zero while (N--) { cout << "X" << num << " = " << Xi; cout << " sin(X" << num << ") = " << fixed; // Find the value of sin() using // inbuilt function cout << setprecision(6) << sin(Xi) << endl; num += 1; // increment by 710 Xi += 710; }}// Driver Codeint main(){ int N = 5; // Function Call printSinX(N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to print all such Xi s.t.// all Xi and sin(Xi) are strictly// increasingstatic void printSinX(int N){ int Xi = 0; int num = 1; // Till N becomes zero while (N-- > 0) { System.out.print("X" + num + " = " + Xi); System.out.print(" sin(X" + num + ") = "); // Find the value of sin() using // inbuilt function System.out.printf("%.6f", Math.sin(Xi)); System.out.println(); num += 1; // Increment by 710 Xi += 710; }}// Driver Codepublic static void main(String[] args){ int N = 5; // Function Call printSinX(N);}}// This code is contributed by Princi Singh |
Python3
# Python3 program for the above approachimport math# Function to print all such Xi s.t.# all Xi and sin(Xi) are strictly# increasingdef printSinX(N): Xi = 0; num = 1; # Till N becomes zero while (N > 0): print("X", num, "=", Xi, end = " "); print("sin(X", num, ") =", end = " "); # Find the value of sin() using # inbuilt function print("{:.6f}".format(math.sin(Xi)), "\n"); num += 1; # increment by 710 Xi += 710; N = N - 1;# Driver CodeN = 5;# Function CallprintSinX(N)# This code is contributed by Code_Mech |
C#
// C# program for the above approachusing System;class GFG{// Function to print all such Xi s.t.// all Xi and sin(Xi) are strictly// increasingstatic void printSinX(int N){ int Xi = 0; int num = 1; // Till N becomes zero while (N-- > 0) { Console.Write("X" + num + " = " + Xi); Console.Write(" sin(X" + num + ") = "); // Find the value of sin() using // inbuilt function Console.Write("{0:F6}", Math.Sin(Xi)); Console.WriteLine(); num += 1; // Increment by 710 Xi += 710; }}// Driver Codepublic static void Main(String[] args){ int N = 5; // Function Call printSinX(N);}}// This code is contributed by SoumikMondal |
Javascript
<script>// Javascript program for the above approach // Function to print all such Xi s.t.// all Xi and sin(Xi) are strictly// increasingfunction printSinX(N){ let Xi = 0; let num = 1; // Till N becomes zero while (N-- > 0) { document.write("X" + num + " = " + Xi); document.write(" sin(X" + num + ") = "); // Find the value of sin() using // inbuilt function document.write(Math.sin(Xi).toFixed(6)); document.write("<br/>"); num += 1; // Increment by 710 Xi += 710; }} // Driver Code let N = 5; // Function Call printSinX(N); </script> |
Output
X1 = 0 sin(X1) = 0.000000 X2 = 710 sin(X2) = 0.000060 X3 = 1420 sin(X3) = 0.000121 X4 = 2130 sin(X4) = 0.000181 X5 = 2840 sin(X5) = 0.000241
Time Complexity: O(N)
Auxiliary Space: O(1)
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