Consider a game where a player can score 3 or 5 or 10 points in a move. Given a total score n, find number of ways to reach the given score.

Examples:

Input: n = 20 Output: 4 There are following 4 ways to reach 20 (10, 10) (5, 5, 10) (5, 5, 5, 5) (3, 3, 3, 3, 3, 5) Input: n = 13 Output: 2 There are following 2 ways to reach 13 (3, 5, 5) (3, 10)

This problem is a variation of coin change problem and can be solved in O(n) time and O(n) auxiliary space.

The idea is to create a table of size n+1 to store counts of all scores from 0 to n. For every possible move (3, 5 and 10), increment values in table.

## C

// A C program to count number of possible ways to a given score // can be reached in a game where a move can earn 3 or 5 or 10 #include <stdio.h> // Returns number of ways to reach score n int count(int n) { // table[i] will store count of solutions for // value i. int table[n+1], i; // Initialize all table values as 0 memset(table, 0, sizeof(table)); // Base case (If given value is 0) table[0] = 1; // One by one consider given 3 moves and update the table[] // values after the index greater than or equal to the // value of the picked move for (i=3; i<=n; i++) table[i] += table[i-3]; for (i=5; i<=n; i++) table[i] += table[i-5]; for (i=10; i<=n; i++) table[i] += table[i-10]; return table[n]; } // Driver program int main(void) { int n = 20; printf("Count for %d is %d\n", n, count(n)); n = 13; printf("Count for %d is %d", n, count(n)); return 0; }

## Java

// Java program to count number of // possible ways to a given score // can be reached in a game where // a move can earn 3 or 5 or 10 import java.util.Arrays; class GFG { // Returns number of ways to reach score n static int count(int n) { // table[i] will store count of solutions for // value i. int table[] = new int[n + 1], i; // Initialize all table values as 0 Arrays.fill(table, 0); // Base case (If given value is 0) table[0] = 1; // One by one consider given 3 // moves and update the table[] // values after the index greater // than or equal to the value of // the picked move for (i = 3; i <= n; i++) table[i] += table[i - 3]; for (i = 5; i <= n; i++) table[i] += table[i - 5]; for (i = 10; i <= n; i++) table[i] += table[i - 10]; return table[n]; } // Driver code public static void main (String[] args) { int n = 20; System.out.println("Count for "+n+" is "+count(n)); n = 13; System.out.println("Count for "+n+" is "+count(n)); } } // This code is contributed by Anant Agarwal.

## Python3

# Python program to count number of possible ways to a given # score can be reached in a game where a move can earn 3 or # 5 or 10. # Returns number of ways to reach score n. def count(n): # table[i] will store count of solutions for value i. # Initialize all table values as 0. table = [0 for i in range(n+1)] # Base case (If given value is 0) table[0] = 1 # One by one consider given 3 moves and update the # table[] values after the index greater than or equal # to the value of the picked move. for i in range(3, n+1): table[i] += table[i-3] for i in range(5, n+1): table[i] += table[i-5] for i in range(10, n+1): table[i] += table[i-10] return table[n] # Driver Program n = 20 print('Count for', n, 'is', count(n)) n = 13 print('Count for', n, 'is', count(n)) # This code is contributed by Soumen Ghosh

## C#

// C# program to count number of // possible ways to a given score // can be reached in a game where // a move can earn 3 or 5 or 10 using System; class GFG { // Returns number of ways to reach // score n static int count(int n) { // table[i] will store count // of solutions for value i. int []table = new int[n + 1]; // Initialize all table values // as 0 for(int j = 0; j < n+1; j++) table[j] = 0; // Base case (If given value is 0) table[0] = 1; // One by one consider given 3 // moves and update the table[] // values after the index greater // than or equal to the value of // the picked move for (int i = 3; i <= n; i++) table[i] += table[i - 3]; for (int i = 5; i <= n; i++) table[i] += table[i - 5]; for (int i = 10; i <= n; i++) table[i] += table[i - 10]; return table[n]; } // Driver code public static void Main () { int n = 20; Console.WriteLine("Count for " + n + " is " + count(n)); n = 13; Console.Write("Count for " + n + " is " + count(n)); } } // This code is contributed by nitin mittal.

Output:

Count for 20 is 4 Count for 13 is 2

**Exercise:** How to count score when (10, 5, 5), (5, 5, 10) and (5, 10, 5) are considered as different sequences of moves. Similarly, (5, 3, 3), (3, 5, 3) and (3, 3, 5) are considered different.

This article is contributed by **Rajeev Arora**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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