POJ 3070 Fibonacci (Matrix Quick Power template)

Description:

In the Fibonacci integer sequence, f0 = 0, f1 = 1, and fn = fn −1 + fn −2 F or n ≥2. For example, the first ten terms of the Fibonacci sequence is:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

An alternative formula for the Fibonacci sequence is

.

Given an integer n, your goal was to compute the last 4 digits of Fn.

Input:

The input test file would contain multiple test cases. Each of the test case consists of a 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 is 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 the 2x2 matrices are given by.

Also, note that raising any 2×2 matrix to the 0th power gives the identity matrix: .

Test instructions: is to require the nth Fibonacci number to 10000 of the remainder, because n is large, simple simulation is certainly not, then led to the Fibonacci of another representation method (the problem has been given), according to the description, we only require the 2*2 matrix {{1,1},{1,0}}^ n can do it.

This leads to a new algorithm: Matrix fast Power (according to the Fast power adaptation, the fast power calculation is the number of the N-square, and this is the matrix of the n-th square).

#include <stdio.h>#include<string.h>#include<queue>#include<math.h>#include<stdlib.h>#include<algorithm>using namespacestd;Const intn=1e6+Ten;Const intinf=0x3f3f3f3f;Const intMod=1e4;typedefLong LongLL;structnode{LL m[2][2];} ANS, TMP, Cnt;node Multiply (Node A, Node B)///calculates the matrix after multiplying two matrices{    intI, j, K;  for(i =0; I <2; i++)    {         for(j =0; J <2; J + +) {Cnt.m[i][j]=0;  for(k =0; K <2; k++) Cnt.m[i][j]= (Cnt.m[i][j]+a.m[i][k]*b.m[k][j])%MOD; }    }    returnCNT;} ll Matrix_power (ll N) {ans.m[0][0] = ans.m[1][1] =1; ans.m[0][1] = ans.m[1][0] =0; tmp.m[0][0] = tmp.m[0][1] = tmp.m[1][0] =1; tmp.m[1][1] =0;  while(n)///and the fast power seeking method is consistent    {        if(n%2!=0) ans=Multiply (ans, TMP); TMP=Multiply (TMP, TMP); N/=2; }    returnans.m[0][1];///since the nth Fibonacci number exists in {fn+1,fn,fn,fn-1}, we return the element at the (0,1) position}intMain () {LL n, num;  while(SCANF ("%lld", &n), n! =-1) {num=matrix_power (n); printf ("%lld\n", num); }    return 0;}

POJ 3070 Fibonacci (Matrix Quick Power template)