# Fibonacci Calculation of large numbers/huge Fibonacci numbers-ACM

Source: Internet
Author: User

Huge Fibonacci numbers

Time Limit: 1 sec memory limit: 128 MB

Description

A Fibonacci sequence is calculated by adding the previous two members
Of the sequence, with the first two members being both 1.

F (1) = 1, F (2) = 1, F (n> 2) = f (n-1) + f (n-2)

Your task is to take a number as input, and print that maid number.

Input
Input one integer.

Output
Output the F (n ).

Notice: No generated Maid number in excess of 1000 digits
Will be in the test data, I. e. F (20) = 6765 has 4 digits.

Sample Input
100

Sample output
354224848179261915075

Key: Addition of large numbers. Nothing else should be done.

I wroteCode:

`# Include <stdio. h> # Include < String . H> /*  A user-defined large number indicates that each int in the method struct array represents a number no more than 10, 0000, and 0000, which is smaller than the maximum value of the signed Int, that is to say, we split a large number into multiple parts for representation, just like when we add two numbers, one bit and one bit, we just useProgram9-bit 9-bit addition, replacing the manual one-bit addition  */  //  The requirement is that the number of 1000-bit Fibonacci is not exceeded. Here I will use max_big BITs, which can represent at most  //  Is max_big * 9 = 120*9 = 1080 bits.  //  Here, big_integer [0] is used to represent the second bit. # Define Max _ big 120 Typedef  Struct  {  //  + 1: it may generate the highest bit carry, but I don't care, directly discard it (otherwise vs2012 will report that the stack is damaged,  //  The stack will indeed be damaged. vs2012 is intelligent) Unsigned Int Big [max_big + 1  ];} Big_integer;  # Define Big_base 1000000000 //  Do not modify it. /*  **************************************** * ************* Function: calc @ 12 function: A + B-> C, large addition parameter: A, B-addition; C: And return: (none) **************************************** **************  */  Void Calc (big_integer * a, big_integer * B, big_integer * C) {unsigned  Int Sum = 0  ;  Int I = 0  ;  /* * ************************ An addition is calculated cyclically below: addend 1: ABCD addend 2: efgh results: (A + E + x) (B + F + x) (C + G + x) (D + H + x) Where: ABCD, efgh here each represents a 9-digit x representing the carry from the low position **************************  */  Memset (C,  0 , Sizeof  (Big_integer ));  For (I = 0 ; I <max_big; I ++ ) {C -> Big [I] = A-> big [I] + B-> big [I] + C-> big [I]; //  Perform 9-bit addition once, remember to add itself C-> big [I +1 ] = C-> big [I]/big_base; //  Add the result carry to the high position C-> big [I] = C-> big [I] % big_base; //  The current bit only retains the remainder  }}  /*  **************************************** * ******************* Function: fibonacci @ 4 function: Calculate the nth Fibonacci value parameter: N: nth digit, the minimum value is 1 ************************************* **********************  */  Void Fibonacci ( Int  N) {big_integer = { 0  }; Big_integer B = { 0  }; Big_integer C = { 0  };  Int I = 0  ;  If (N <= 1  ) {Printf (  "  % D \ n  " , N =1  );  Return  ;}  //  Fibonacci ,...  //  To guarantee the results of the first calculation (n = 2), the initial values are: a = 0, B = 1 => C = 1 A. Big [ 0 ] = 0  ; B. Big [  0 ] = 1  ;  /* * ************************ The following cyclically calculates the onacci like decimal places: F3 = F1 + F2; f1 = F2; F2 = F3 ;**************************  */      For (I = 2 ; I <= N; I ++ ) {Calc ( & A, & B ,& C); memcpy ( & A, & B, Sizeof  (Big_integer); memcpy ( & B, & C, Sizeof  (Big_integer ));}  // This for loop ignores a numerical bit (9 digits) whose value starts from the highest bit)  //  Of course, condition I> 0 is required. Do you know why?      For (I = max_big- 1 ; C. Big [I] = 0 & I> 0 ; I -- );  //  For the highest bit: The leading '0' is not required' Printf ( "  % U  " , C. Big [I -- ]); //  Then, each numeric bit is output in 9-bit decomposition. Remember to output leading 0,  //  At first, I didn't notice it.      While (I> = 0  ) {Printf (  "  % 09u  "  , C. Big [I]); I -- ;} Printf (  "  \ N  "  );} Int Main ( Void  ){  Int  N; scanf (  "  % D  " ,& N); maid (N );  Return   0  ;} `

The program runs at a very fast speed. In fact, it takes less than 1 second to calculate to 10000:

Girl don't cry @ 01:36:13 @ http://www.cnblogs.com/nbsofer

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