## 斐波那契数列

0、1、1、2、3、5、8、13、21、34......

F(1)=1，F(2)=1,
F(n)=F(n-1)+F(n-2)（n ≥ 3，n ∈ N*）

## 求第n项斐波那契数

``````//Fibonacci Series using Recursion
#include<stdio.h>
int fib(int n) {
if (n <= 1)
return n;
return fib(n - 1) + fib(n - 2);
}

int main() {
int n = 9;
printf("%d", fib(n));

return 0;
}``````

《深入浅出理解动态规划（一） | 交叠子问题》

《深入浅出理解动态规划（二） | 最优子结构》

``````//Fibonacci Series using Dynamic Programming
#include<stdio.h>

int fib(int n) {
/* Declare an array to store Fibonacci numbers. */
int f[n + 1];
int i;

/* 0th and 1st number of the series are 0 and 1*/
f[0] = 0;
f[1] = 1;

for (i = 2; i <= n; i++) {
/* Add the previous 2 numbers in the series
and store it */
f[i] = f[i - 1] + f[i - 2];
}

return f[n];
}

int main() {
int n = 9;
printf("%d", fib(n));

return 0;
}``````

``````// Fibonacci Series using Space Optimized Method
#include<stdio.h>
int fib(int n) {
int a = 0, b = 1, c, i;
if (n == 0)
return a;
for (i = 2; i <= n; i++) {
c = a + b;
a = b;
b = c;
}
return b;
}

int main() {
int n = 9;
printf("%d", fib(n));

return 0;
}``````

``````//Fibonacci Series using Dynamic Programming
#include<stdio.h>

int fib(int n) {
/* Declare an array to store Fibonacci numbers. */
int f[3];       /* 只需定义一个长度为3的数组 */
int i;

/* 0th and 1st number of the series are 0 and 1*/
f[0] = 0;
f[1] = 1;

for (i = 2; i <= n; i++) {
/* Add the previous 2 numbers in the series
and store it：注意下标要对3取模 */
f[i % 3] = f[(i - 1) % 3] + f[(i - 2) % 3];
}

return f[n % 3]; /* 这里要注意下标对3取模 */
}

int main() {
int n = 9;
printf("%d", fib(n));

return 0;
}
``````

``````#include <stdio.h>

/* Helper function that multiplies 2 matrices F and M of size 2*2, and puts the multiplication result back to F[][] */
void multiply(int F[2][2], int M[2][2]);

/* Helper function that calculates F[][] raise to the power n and puts the result in F[][]。Note that this function is designed only for fib() and won't work as general power function */
void power(int F[2][2], int n);

int fib(int n) {
int F[2][2] = { { 1, 1 }, { 1, 0 } };
if (n == 0)
return 0;
power(F, n - 1);

return F[0][0];
}

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;
}

void power(int F[2][2], int n) {
int i;
int M[2][2] = { { 1, 1 }, { 1, 0 } };

// n - 1 times multiply the matrix to {{1,0},{0,1}}
for (i = 2; i <= n; i++)
multiply(F, M);
}

/* Driver program to test above function */
int main() {
int n = 9;
printf("%d", fib(n));

return 0;
}``````

``````#include <stdio.h>

void multiply(int F[2][2], int M[2][2]);

void power(int F[2][2], int n);

/* function that returns nth Fibonacci number */
int fib(int n) {
int F[2][2] = { { 1, 1 }, { 1, 0 } };
if (n == 0)
return 0;
power(F, n - 1);
return F[0][0];
}

/* Optimized version of power() in method 4 */
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);
}

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;
}

/* Driver program to test above function */
int main() {
int n = 9;
printf("%d", fib(n));

return 0;
}``````

F(n)=[2F(k-1)+F(k)]F(k)

F(n)=F(k)F(k)+F(k-1)F(k-1)

#### 原文链接：原文来自个人公众号：C you again，欢迎关注

``````// C++ Program to find n'th fibonacci Number in
// with O(Log n) arithmatic operations
#include <bits/stdc++.h>
using namespace std;

const int MAX = 1000;

// Create an array for memoization
int f[MAX] = { 0 };

// Returns n'th fuibonacci number using table f[]
int fib(int n) {
// Base cases
if (n == 0)
return 0;
if (n == 1 || n == 2)
return (f[n] = 1);

// If fib(n) is already computed
if (f[n])
return f[n];

int k = (n & 1) ? (n + 1) / 2 : n / 2;

// Applyting above formula [Note value n&1 is 1
// if n is odd, else 0.
f[n] = (n & 1) ?
(fib(k) * fib(k) + fib(k - 1) * fib(k - 1)) :
(2 * fib(k - 1) + fib(k)) * fib(k);

return f[n];
}

/* Driver program to test above function */
int main() {
int n = 9;
printf("%d ", fib(n));
return 0;
}
``````

Java提供了BigInteger类，可以很轻易地算出当n很大时的斐波那契数。

``````// Java program to compute n-th Fibonacci number where n may be large.
import java.math.*;

public class Fibonacci {
// Returns n-th Fibonacci number
static BigInteger fib(int n) {
BigInteger a = BigInteger.valueOf(0);
BigInteger b = BigInteger.valueOf(1);
BigInteger c = BigInteger.valueOf(1);
for (int j = 2; j <= n; j++) {
a = b;
b = c;
}

return (a);
}

public static void main(String[] args) {
int n = 1000;
System.out.println("Fibonacci of " + n + "th term" + " " + "is" + " " + fib(n));
}
}
``````

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