1 #include <stdio.h>
 2 #include <stdlib.h>
 3
 4 long long Fibonacci(int n)
 5 {
 6 long long f;
 7 if(n==1||n==2)
 8 f=1;
 9 else
10 f=Fibonacci(n-1)+Fibonacci(n-2);
11 return f;
12 }
13
14 int main()
15 {
16 int n;
17 long long f;
18 scanf("%d",&n);
19 f=Fibonacci(n);
20 printf("%lld",f);
21 return 0;
22 }

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3
 4 long long Fibonacci(int n)
 5 {
 6 int result[2]={0,1},i;
 7 if(n<2)
 8 return result[n];
 9 long long One = 1,Two = 0,Current_N = 0;
10 for(i=2;i<=n;i++)
11 {
12 Current_N = One + Two;
13 Two = One;
14 One = Current_N;
15 }
16 return Current_N;
17 }
18
19 int main()
20 {
21 long long n,f;
22 scanf("%d",&n);
23 f=Fibonacci(n);
24 printf("%lld",f);
25 return 0;
26 }

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstdlib>
 4
 5 using namespace std;
 6
 7 //求矩阵a的n次幂的函数
 8 long long * Matrix(long long *a,int n)
 9 {
10 long long *result = (long long *)malloc(sizeof(long long)*4);
11 long long *finnal = (long long *)malloc(sizeof(long long)*4);
12 //如果n等于1的话，则直接返回矩阵a，毕竟，一次幂就不必求了。
13 if(n==1)
14 return a;
15 //先求出矩阵的n/2次幂
16 long long *CurMatrix = Matrix(a,n/2);
17 //两个矩阵的n/2次幂相乘得到矩阵的n次幂
18 result[0] = CurMatrix[0]*CurMatrix[0]+CurMatrix[1]*CurMatrix[2];
19 result[1] = CurMatrix[0]*CurMatrix[1]+CurMatrix[1]*CurMatrix[3];
20 result[2] = CurMatrix[2]*CurMatrix[0]+CurMatrix[3]*CurMatrix[2];
21 result[3] = CurMatrix[2]*CurMatrix[1]+CurMatrix[3]*CurMatrix[3];
22 //如果n为奇数的话，则(n/2)会少一位，补回来。a = a^(n/2)*a^(n/2) * a;
23 if(n%2==1)
24 {
25 finnal[0] = result[0]*a[0]+result[1]*a[2];
26 finnal[1] = result[0]*a[1]+result[1]*a[3];
27 finnal[2] = result[2]*a[0]+result[3]*a[2];
28 finnal[3] = result[2]*a[1]+result[3]*a[3];
29 }
30 //如果n为偶数的话，n/2还是n/2
31 else
32 finnal = result;
33 return finnal;
34 }
35
36 int main(void)
37 {
38 long long a[4];
39 int n;
40 cin>>n;
41 //矩阵按一位数组设置,t[0]即为febonacci(n)
42 a[0] = 1;
43 a[1] = 1;
44 a[2] = 1;
45 a[3] = 0;
46 long long *t = Matrix(a,n-1);
47 cout<<t[0]<<endl;
48 return 0;
49 }