poj 3070 Fibonacci（矩阵加速DP）

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Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12459 Accepted: 8853

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_n_{− 1} + F_n_{− 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 F_n.

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 F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n 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:

Source

Stanford Local 2006

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题解：矩阵加速DP。

学到了一个很高效的求解fibonacci的方法。

1 1 fn+1 fn 1 1 fn+1+fn fn+1

1 0 这个矩阵既是加速矩阵也是初始矩阵。 fn fn-1 × 1 0 =fn+fn+1 fn

所以要求序列第K项的答案就是求矩阵A^k 的 a[1][2]

 #include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#define N 2#define p 10000using namespace std;int n,m;struct data {int a[10][10];}a;void clear(data &x){for (int i=1;i<=N;i++) for (int j=1;j<=N;j++)  x.a[i][j]=0;}void change(data &a,data b){for (int i=1;i<=N;i++) for (int j=1;j<=N;j++)  a.a[i][j]=b.a[i][j];}data mul(data a,data b){data c;for (int i=1;i<=N;i++) for (int j=1;j<=N;j++)  {  c.a[i][j]=0;  for (int k=1;k<=N;k++)   c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j]%p)%p;  }return c;}data pow(data num,int x){data ans; clear(ans);for (int i=1;i<=N;i++) ans.a[i][i]=1;data base; change(base,num);  while (x){if (x&1)  ans=mul(ans,base);x>>=1;base=mul(base,base);}return ans;}int main(){freopen("a.in","r",stdin);a.a[1][1]=1; a.a[1][2]=1; a.a[2][1]=1; a.a[2][2]=0;while (scanf("%d",&n)!=EOF){if (n==-1) break;if (n==0){printf("0\n");continue;}data ans=pow(a,n);printf("%d\n",ans.a[1][2]);}}

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