POJ——3070快速幂

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, andF_n = F_{n − 1} + F_{n − 2} forn ≥ 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 ofF_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., printF_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:

.

基本的快速幂的方法：代码易懂

#include <iostream> #include <stdio.h> using namespace std; struct Mat{    int num[3][3]; }; Mat mult(Mat a,Mat b){    Mat c;    for(int i=1;i<=2;i++)      for(int j=1;j<=2;j++){         c.num[i][j]=0;        for(int k=1;k<=2;k++)          c.num[i][j]=(c.num[i][j]+a.num[i][k]*b.num[k][j])%10000;      }   return c; } Mat pow(Mat x,int y){    Mat ans;    for(int i=1;i<=2;i++)     for(int j=1;j<=2;j++){         if(i==j)            ans.num[i][j]=1;         else            ans.num[i][j]=0;     }   while(y){        if(y&1)          ans=mult(ans,x);          y>>=1;        x=mult(x,x);    }  return ans; } int main(){     int n;   while(scanf("%d",&n)){      if(n==-1)         break;      Mat A;      A.num[1][1]=1;      A.num[1][2]=1;      A.num[2][1]=1;      A.num[2][2]=0;      A=pow(A,n);    printf("%d\n",A.num[1][2]);   }   }