Given two positive integers N and M, the task is to find the number of ways to place M distinct objects in partitions of even indexed boxes which are numbered [1, N] sequentially, and every ith Box has i distinct partitions. Since the answer can be very large, print modulo 1000000007.
Examples:
Input: N = 2, M = 1
Output: 2
Explanation: Since, N = 2. There is only one even indexed box i.e box 2, having 2 partitions. Therefore, there are two positions to place an object. Therefore, number of ways = 2.Input: N = 5, M = 2
Output: 32
Approach: Follow the steps below to solve the problem:
- M objects are to be placed in even indexed box’s partitions. Let S be the total even indexed box’s partitions in N boxes.
- The number of partitions is equal to the summation of all even numbers up to N. Therefore, Sum, S = X * (X + 1), where X = floor(N / 2).
- Each object can occupy one of S different positions. Therefore, the total number of ways = S*S*S..(M times) = SM.
Below is the implementation of the above approach:
C++
// C++ implementation of the// above Approach#include <bits/stdc++.h>using namespace std;const int MOD = 1000000007;// Iterative Function to calculate// (x^y)%p in O(log y)int power(int x, unsigned int y, int p = MOD){ // Initialize Result int res = 1; // Update x if x >= MOD // to avoid multiplication overflow x = x % p; while (y > 0) { // If y is odd, multiply x with result if (y & 1) res = (res * 1LL * x) % p; // multiplied by long long int, // to avoid overflow // because res * x <= 1e18, which is // out of bounds for integer // n must be even now // y = y/2 y = y >> 1; // Change x to x^2 x = (x * 1LL * x) % p; } return res;}// Utility function to find// the Total Number of Waysvoid totalWays(int N, int M){ // Number of Even Indexed // Boxes int X = N / 2; // Number of partitions of // Even Indexed Boxes int S = (X * 1LL * (X + 1)) % MOD; // Number of ways to distribute // objects cout << power(S, M, MOD) << "\n";}// Driver Codeint main(){ // N = number of boxes // M = number of distinct objects int N = 5, M = 2; // Function call to // get Total Number of Ways totalWays(N, M); return 0;} |
Java
// Java implementation of the// above Approachimport java.io.*;class GFG{ public static int MOD = 1000000007;// Iterative Function to calculate// (x^y)%p in O(log y)static int power(int x, int y, int p){ p = MOD; // Initialize Result int res = 1; // Update x if x >= MOD // to avoid multiplication overflow x = x % p; while (y > 0) { // If y is odd, multiply x with result if ((y & 1) != 0) res = (res * x) % p; // multiplied by long long int, // to avoid overflow // because res * x <= 1e18, which is // out of bounds for integer // n must be even now // y = y/2 y = y >> 1; // Change x to x^2 x = (x * x) % p; } return res;}// Utility function to find// the Total Number of Waysstatic void totalWays(int N, int M){ // Number of Even Indexed // Boxes int X = N / 2; // Number of partitions of // Even Indexed Boxes int S = (X * (X + 1)) % MOD; // Number of ways to distribute // objects System.out.println(power(S, M, MOD));}// Driver Codepublic static void main (String[] args) { // N = number of boxes // M = number of distinct objects int N = 5, M = 2; // Function call to // get Total Number of Ways totalWays(N, M);}}// This code is contributed by Dharanendra L V |
Python3
# Python3 implementation of the# above ApproachMOD = 1000000007# Iterative Function to calculate# (x^y)%p in O(log y)def power(x, y, p = MOD): # Initialize Result res = 1 # Update x if x >= MOD # to avoid multiplication overflow x = x % p while (y > 0): # If y is odd, multiply x with result if (y & 1): res = (res * x) % p # multiplied by long long int, # to avoid overflow # because res * x <= 1e18, which is # out of bounds for integer # n must be even now # y = y/2 y = y >> 1 # Change x to x^2 x = (x * x) % p return res# Utility function to find# the Total Number of Waysdef totalWays(N, M): # Number of Even Indexed # Boxes X = N // 2 # Number of partitions of # Even Indexed Boxes S = (X * (X + 1)) % MOD # Number of ways to distribute # objects print (power(S, M, MOD))# Driver Codeif __name__ == '__main__': # N = number of boxes # M = number of distinct objects N, M = 5, 2 # Function call to # get Total Number of Ways totalWays(N, M)# This code is contributed by mohit kumar 29. |
C#
// C# implementation of the// above Approachusing System;public class GFG{public static int MOD = 1000000007;// Iterative Function to calculate// (x^y)%p in O(log y)static int power(int x, int y, int p){ p = MOD; // Initialize Result int res = 1; // Update x if x >= MOD // to avoid multiplication overflow x = x % p; while (y > 0) { // If y is odd, multiply x with result if ((y & 1) != 0) res = (res * x) % p; // multiplied by long long int, // to avoid overflow // because res * x <= 1e18, which is // out of bounds for integer // n must be even now // y = y/2 y = y >> 1; // Change x to x^2 x = (x * x) % p; } return res;}// Utility function to find// the Total Number of Waysstatic void totalWays(int N, int M){ // Number of Even Indexed // Boxes int X = N / 2; // Number of partitions of // Even Indexed Boxes int S = (X * (X + 1)) % MOD; // Number of ways to distribute // objects Console.WriteLine(power(S, M, MOD));}// Driver Codestatic public void Main (){ // N = number of boxes // M = number of distinct objects int N = 5, M = 2; // Function call to // get Total Number of Ways totalWays(N, M);}}// This code is contributed by Dharanendra L V |
Javascript
<script>// Javascript implementation of the// above Approachvar MOD = 1000000007;// Iterative Function to calculate// (x^y)%p in O(log y)function power(x, y, p = MOD){ // Initialize Result var res = 1; // Update x if x >= MOD // to avoid multiplication overflow x = x % p; while (y > 0) { // If y is odd, multiply x with result if (y & 1) res = (res * 1 * x) % p; // multiplied by long long int, // to avoid overflow // because res * x <= 1e18, which is // out of bounds for integer // n must be even now // y = y/2 y = y >> 1; // Change x to x^2 x = (x * 1 * x) % p; } return res;}// Utility function to find// the Total Number of Waysfunction totalWays(N, M){ // Number of Even Indexed // Boxes var X = parseInt(N / 2); // Number of partitions of // Even Indexed Boxes var S = (X * 1 * (X + 1)) % MOD; // Number of ways to distribute // objects document.write( power(S, M, MOD) << "<br>");}// Driver Code// N = number of boxes// M = number of distinct objectsvar N = 5, M = 2;// Function call to// get Total Number of WaystotalWays(N, M);</script> |
Output:
36
Time Complexity: O(log M)
Auxiliary Space: O(1)
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