# Ways to represent a number as a sum of 1’s and 2’s

Given a positive integer **N**. The task is to find the number of ways of representing N as a sum of 1s and 2s.**Examples:**

Input : N = 3 Output : 3 3 can be represented as (1+1+1), (2+1), (1+2). Input : N = 5 Output : 8

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For N = 1, answer is 1.

For N = 2. (1 + 1), (2), answer is 2.

For N = 3. (1 + 1 + 1), (2 + 1), (1 + 2), answer is 3.

For N = 4. (1 + 1 + 1 + 1), (2 + 1 + 1), (1 + 2 + 1), (1 + 1 + 2), (2 + 2) answer is 5.

And so on.

It can be observed that it form Fibonacci Series. So, the number of ways of representing N as a sum of 1s and 2s is (N + 1)^{th} Fibonacci number. *How?*

We can easily see that the recursive function is exactly the same as Fibonacci Numbers. To obtain the sum of N, we can add 1 to N – 1. Also, we can add 2 to N – 2. And only 1 and 2 are allowed to make the sum N. So, to obtain sum N using 1s and 2s, total ways are: number of ways to obtain (N – 1) + number of ways to obtain (N – 2).

We can find N’th Fibonacci Number in O(Log n) time. Please refer to method 5 of this post.

Below is the implementation of this approach:

## C++

`// C++ program to find number of ways to representing` `// a number as a sum of 1's and 2's` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to multiply matrix.` `void` `multiply(` `int` `F[2][2], ` `int` `M[2][2])` `{` ` ` `int` `x = F[0][0]*M[0][0] + F[0][1]*M[1][0];` ` ` `int` `y = F[0][0]*M[0][1] + F[0][1]*M[1][1];` ` ` `int` `z = F[1][0]*M[0][0] + F[1][1]*M[1][0];` ` ` `int` `w = F[1][0]*M[0][1] + F[1][1]*M[1][1];` ` ` `F[0][0] = x;` ` ` `F[0][1] = y;` ` ` `F[1][0] = z;` ` ` `F[1][1] = w;` `}` `// Power function in log n` `void` `power(` `int` `F[2][2], ` `int` `n)` `{` ` ` `if` `( n == 0 || n == 1)` ` ` `return` `;` ` ` `int` `M[2][2] = {{1,1},{1,0}};` ` ` `power(F, n/2);` ` ` `multiply(F, F);` ` ` `if` `(n%2 != 0)` ` ` `multiply(F, M);` `}` `/* function that returns (n+1)th Fibonacci number` ` ` `Or number of ways to represent n as sum of 1's` ` ` `2's */` `int` `countWays(` `int` `n)` `{` ` ` `int` `F[2][2] = {{1,1},{1,0}};` ` ` `if` `(n == 0)` ` ` `return` `0;` ` ` `power(F, n);` ` ` `return` `F[0][0];` `}` `// Driver program` `int` `main()` `{` ` ` `int` `n = 5;` ` ` `cout << countWays(n) << endl;` ` ` `return` `0;` `}` |

## Java

`// Java program to find number of` `// ways to representing a number` `// as a sum of 1's and 2's` `class` `GFG` `{` `// Function to multiply matrix.` ` ` `static` `void` `multiply(` `int` `F[][], ` `int` `M[][])` ` ` `{` ` ` `int` `x = F[` `0` `][` `0` `] * M[` `0` `][` `0` `] + F[` `0` `][` `1` `] * M[` `1` `][` `0` `];` ` ` `int` `y = F[` `0` `][` `0` `] * M[` `0` `][` `1` `] + F[` `0` `][` `1` `] * M[` `1` `][` `1` `];` ` ` `int` `z = F[` `1` `][` `0` `] * M[` `0` `][` `0` `] + F[` `1` `][` `1` `] * M[` `1` `][` `0` `];` ` ` `int` `w = F[` `1` `][` `0` `] * M[` `0` `][` `1` `] + F[` `1` `][` `1` `] * M[` `1` `][` `1` `];` ` ` `F[` `0` `][` `0` `] = x;` ` ` `F[` `0` `][` `1` `] = y;` ` ` `F[` `1` `][` `0` `] = z;` ` ` `F[` `1` `][` `1` `] = w;` ` ` `}` ` ` `// Power function in log n` ` ` `static` `void` `power(` `int` `F[][], ` `int` `n)` ` ` `{` ` ` `if` `(n == ` `0` `|| n == ` `1` `)` ` ` `{` ` ` `return` `;` ` ` `}` ` ` `int` `M[][] = {{` `1` `, ` `1` `}, {` `1` `, ` `0` `}};` ` ` `power(F, n / ` `2` `);` ` ` `multiply(F, F);` ` ` `if` `(n % ` `2` `!= ` `0` `)` ` ` `{` ` ` `multiply(F, M);` ` ` `}` ` ` `}` ` ` `/* function that returns (n+1)th Fibonacci number` ` ` `Or number of ways to represent n as sum of 1's` ` ` `2's */` ` ` `static` `int` `countWays(` `int` `n)` ` ` `{` ` ` `int` `F[][] = {{` `1` `, ` `1` `}, {` `1` `, ` `0` `}};` ` ` `if` `(n == ` `0` `)` ` ` `{` ` ` `return` `0` `;` ` ` `}` ` ` `power(F, n);` ` ` `return` `F[` `0` `][` `0` `];` ` ` `}` ` ` `// Driver program` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `n = ` `5` `;` ` ` `System.out.println(countWays(n));` ` ` `}` `}` `// This code contributed by Rajput-Ji` |

## Python3

`# Python3 program to find number of ways to` `# representing a number as a sum of 1's and 2's` `# Function to multiply matrix.` `def` `multiply(F, M):` ` ` `x ` `=` `F[` `0` `][` `0` `] ` `*` `M[` `0` `][` `0` `] ` `+` `F[` `0` `][` `1` `] ` `*` `M[` `1` `][` `0` `]` ` ` `y ` `=` `F[` `0` `][` `0` `] ` `*` `M[` `0` `][` `1` `] ` `+` `F[` `0` `][` `1` `] ` `*` `M[` `1` `][` `1` `]` ` ` `z ` `=` `F[` `1` `][` `0` `] ` `*` `M[` `0` `][` `0` `] ` `+` `F[` `1` `][` `1` `] ` `*` `M[` `1` `][` `0` `]` ` ` `w ` `=` `F[` `1` `][` `0` `] ` `*` `M[` `0` `][` `1` `] ` `+` `F[` `1` `][` `1` `] ` `*` `M[` `1` `][` `1` `]` ` ` `F[` `0` `][` `0` `] ` `=` `x` ` ` `F[` `0` `][` `1` `] ` `=` `y` ` ` `F[` `1` `][` `0` `] ` `=` `z` ` ` `F[` `1` `][` `1` `] ` `=` `w` `# Power function in log n` `def` `power(F, n):` ` ` `if` `( n ` `=` `=` `0` `or` `n ` `=` `=` `1` `):` ` ` `return` ` ` `M ` `=` `[[` `1` `, ` `1` `],[` `1` `, ` `0` `]]` ` ` `power(F, n ` `/` `/` `2` `)` ` ` `multiply(F, F)` ` ` `if` `(n ` `%` `2` `!` `=` `0` `):` ` ` `multiply(F, M)` `#/* function that returns (n+1)th Fibonacci number` `# Or number of ways to represent n as sum of 1's` `# 2's */` `def` `countWays(n):` ` ` `F ` `=` `[[` `1` `, ` `1` `], [` `1` `, ` `0` `]]` ` ` `if` `(n ` `=` `=` `0` `):` ` ` `return` `0` ` ` `power(F, n)` ` ` `return` `F[` `0` `][` `0` `]` `# Driver Code` `n ` `=` `5` `print` `(countWays(n))` `# This code is contributed by mohit kumar` |

## C#

`// C# program to find number of` `// ways to representing a number` `// as a sum of 1's and 2's` `class` `GFG` `{` ` ` `// Function to multiply matrix.` ` ` `static` `void` `multiply(` `int` `[,]F, ` `int` `[,]M)` ` ` `{` ` ` `int` `x = F[0,0] * M[0,0] + F[0,1] * M[1,0];` ` ` `int` `y = F[0,0] * M[0,1] + F[0,1] * M[1,1];` ` ` `int` `z = F[1,0] * M[0,0] + F[1,1] * M[1,0];` ` ` `int` `w = F[1,0] * M[0,1] + F[1,1] * M[1,1];` ` ` `F[0,0] = x;` ` ` `F[0,1] = y;` ` ` `F[1,0] = z;` ` ` `F[1,1] = w;` ` ` `}` ` ` `// Power function in log n` ` ` `static` `void` `power(` `int` `[,]F, ` `int` `n)` ` ` `{` ` ` `if` `(n == 0 || n == 1)` ` ` `{` ` ` `return` `;` ` ` `}` ` ` `int` `[,]M = {{1, 1}, {1, 0}};` ` ` `power(F, n / 2);` ` ` `multiply(F, F);` ` ` `if` `(n % 2 != 0)` ` ` `{` ` ` `multiply(F, M);` ` ` `}` ` ` `}` ` ` `/* function that returns (n+1)th Fibonacci number` ` ` `Or number of ways to represent n as sum of 1's` ` ` `2's */` ` ` `static` `int` `countWays(` `int` `n)` ` ` `{` ` ` `int` `[,]F = {{1, 1}, {1, 0}};` ` ` `if` `(n == 0)` ` ` `{` ` ` `return` `0;` ` ` `}` ` ` `power(F, n);` ` ` `return` `F[0,0];` ` ` `}` ` ` `// Driver program` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `n = 5;` ` ` `System.Console.WriteLine(countWays(n));` ` ` `}` `}` `// This code contributed by mits` |

## Javascript

`<script>` `// Javascript program to find number of` `// ways to representing a number` `// as a sum of 1's and 2's` `// Function to multiply matrix.` `function` `multiply(F , M)` `{` ` ` `var` `x = F[0][0] * M[0][0] + F[0][1] * M[1][0];` ` ` `var` `y = F[0][0] * M[0][1] + F[0][1] * M[1][1];` ` ` `var` `z = F[1][0] * M[0][0] + F[1][1] * M[1][0];` ` ` `var` `w = F[1][0] * M[0][1] + F[1][1] * M[1][1];` ` ` `F[0][0] = x;` ` ` `F[0][1] = y;` ` ` `F[1][0] = z;` ` ` `F[1][1] = w;` `}` `// Power function in log n` `function` `power(F , n)` `{` ` ` `if` `(n == 0 || n == 1)` ` ` `{` ` ` `return` `;` ` ` `}` ` ` `var` `M = [[1, 1], [1, 0]];` ` ` `power(F, parseInt(n / 2));` ` ` `multiply(F, F);` ` ` `if` `(n % 2 != 0)` ` ` `{` ` ` `multiply(F, M);` ` ` `}` `}` `/* function that returns (n+1)th Fibonacci number` `Or number of ways to represent n as sum of 1's` `2's */` `function` `countWays(n)` `{` ` ` `var` `F = [[1, 1], [1, 0]];` ` ` `if` `(n == 0)` ` ` `{` ` ` `return` `0;` ` ` `}` ` ` `power(F, n);` ` ` `return` `F[0][0];` `}` `// Driver program` `var` `n = 5;` `document.write(countWays(n));` `// This code is contributed by 29AjayKumar` `</script>` |

**Output:**

8

**Time Complexity: **O(logn).

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