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poj3070 (斐波那契,矩阵快速幂)
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
099999999991000000000-1
Sample Output
0346266875
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 matrix13 {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 else58 {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 }
poj3070 (斐波那契,矩阵快速幂)