POJ 3070 Fibonacci "matrix fast power to find the nth Fibonacci number%1000"

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11123		Accepted: 7913

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 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 A/2x2 matrices is given by

.

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

.

Test instructions: give you a formula to solve the nth Fibonacci number. Let you find the FN% 10000.

Constructs the unit matrix, then is the Matrix fast power.

AC Code:

#include <cstdio> #include <cstring> #include <algorithm> #define MAXN 100#define LL Long Long#define    MOD 10000using namespace Std;struct matrix{LL A[MAXN][MAXN]; int R, c;//number of rows};    Matrix Ori, res;//initial matrix and result matrix void init ()//initialization matrix {memset (RES.A, 0, sizeof (RES.A)); RES.R = 2;    RES.C = 2;    for (int i = 1; I <= 2; i++)//tectonic unit matrix res.a[i][i] = 1; ORI.R = 2;    ORI.C = 2;    ORI.A[1][1] = ori.a[1][2] = ori.a[2][1] = 1; ORI.A[2][2] = 0;}    Matrix multi (Matrix x, Matrix y) {matrix z;    memset (z.a, 0, sizeof (Z.A)); Z.R = X.R, z.c = y.c;//the number of new matrix rows equals the number of rows of the x matrix equals the number of columns of the y matrix for (int i = 1; I <= X.R; i++) The number of rows of the//x matrix {for (int k = 1; K & Lt;= x.c;  k++)//matrix x the number of columns equal to the number of rows of the matrix Y is x.c = y.r {if (x.a[i][k] = = 0) continue;//optimization for (int j = 1; j<= y.c;        J + +)//y matrix number of columns z.a[i][j] = (Z.a[i][j] + (x.a[i][k] * y.a[k][j])% mod)% MoD; }} return z;}       void Matrix_mod (int n) {while (n)//n Power {if (N & 1)     res = multi (ori, RES);        Ori = multi (ori, ORI);    n >>= 1; } printf ("%lld\n", Res.a[1][2]% MOD);}    int main () {int N; while (scanf ("%d", &n), n!=-1) {init ();//Initialize the unit matrix Matrix_mod (N);//Matrix Fast Power} return 0;}

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

POJ 3070 Fibonacci "matrix fast power to find the nth Fibonacci number%1000"