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

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 observe 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 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 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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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**Output:**

8

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

This article is contributed by **Anuj Chauhan**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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