Fibonacci
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9630 | Accepted: 6839 |
Description
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
An alternative formula for the Fibonacci sequence is
.
Given an integer n, your goal is to compute the last 4 digits of Fn.
Input
The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.
Output
For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
Sample Input
0 9 999999999 1000000000 -1
Sample Output
0 34 626 6875
Hint
As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by
.
Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:
.
求斐波那契序列的公式。
由于该矩阵的特殊结构使得a(n+1)[0][0] = a(n)[0][0]+a(n)[0][1], a(n+1)[0][1] = a(n)[1][1], a(n+1)[1][0] = a(n)[0][1]+a(n)[1][0], a(n+1)[1][1] = a(n)[1][0];
code:
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<vector>
5 #include<algorithm>
6 #include<cmath>
7 #define M(a,b) memset(a,b,sizeof(a))
8
9 using namespace std;
10
11 int n;
12 struct matrix
13 {
14 int a[2][2];
15 void init()
16 {
17 a[0][0] = a[1][0] = a[0][1] = 1;
18 a[1][1] = 0;
19 }
20 };
21
22 matrix mamul(matrix a,matrix b)
23 {
24 matrix c;
25 for(int i = 0;i<2;i++)
26 {
27 for(int j = 0;j<2;j++)
28 {
29 c.a[i][j] = 0;
30 for(int k = 0;k<2;k++)
31 c.a[i][j]+=(a.a[i][k]*b.a[k][j]);
32 c.a[i][j]%=10000;
33 }
34 }
35 return c;
36 }
37
38 matrix mul(matrix s, int k)
39 {
40 matrix ans;
41 ans.init();
42 while(k>=1)
43 {
44 if(k&1)
45 ans = mamul(ans,s);
46 k = k>>1;
47 s = mamul(s,s);
48 }
49 return ans;
50 }
51
52 int main()
53 {
54 while(scanf("%d",&n)==1&n>=0)
55 {
56 if(n==0) puts("0");
57 else
58 {
59 matrix ans;
60 ans.init();
61 ans = mul(ans,n-1);
62 printf("%d\n",ans.a[0][1]%10000);
63 }
64 }
65 return 0;
66 }
下面代码只是测试公式，无法解决取模的问题，因为中间为double型，无法取模：
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<vector>
5 #include<algorithm>
6 #include<cmath>
7 #define M(a,b) memset(a,b,sizeof(a))
8
9 using namespace std;
10
11 double Pow(double a,int n)
12 {
13 double ans = 1;
14 while(n>=1)
15 {
16 if(n&1)
17 ans = a*ans;
18 n = n>>1;
19 a = a*a;
20 }
21 return ans;
22 }
23
24 int main()
25 {
26 int n;
27 double a = (sqrt(5.0)+1.0)/2;
28 double b = (-sqrt(5.0)+1.0)/2;
29 double c = (sqrt(5.0))/5;
30 while(scanf("%d",&n)==1)
31 {
32 int ans = (int)(c*(Pow(a,n)-Pow(b,n)))%10000;
33 printf("%d\n",ans);
34 }
35 return 0;
36 }