Poj 3070 Fibonacci [matrix concatenation]

Fibonacci

Time limit:1000 ms		Memory limit:65536 K
Total submissions:6881		Accepted:4873

Description

In the Fibonacci integer sequence,F0 = 0,F1 = 1, andFN=FN− 1 +FN−2N≥2. For example, the first ten terms
The maid sequence are:

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

An alternative formula for the Fibonacci sequence is

.

Given an integerN, Your goal is to compute the last 4 digitsFN.

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 digitsFN. If the last four digitsFNAre all zeros, print '0'; otherwise, omit any leading zeros (I. e., printFNMoD 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

.

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

.

Source

Stanford local 2006

3070

Accepted

132 k

0 ms

C ++

20172b

21:17:22

#include<cstdio>const int mod = 10000;struct Matrix{    __int64 a11, a12, a21, a22;}matrix;Matrix mull(Matrix a, Matrix b){    Matrix c;    c.a11 = a.a11*b.a11+a.a12*b.a21;    c.a12 = a.a11*b.a12+a.a12*b.a22;    c.a21 = a.a21*b.a11+a.a22*b.a21;    c.a22 = a.a21*b.a12+a.a22*b.a22;    c.a11 %= mod;    c.a12 %= mod;    c.a21 %= mod;    c.a22 %= mod;    return c;}Matrix find(Matrix m, __int64 n){    Matrix b;    b.a11 = 1; b.a12 = 0;    b.a21 = 0; b.a22 = 1;    while(n > 0)    {        if(n&1)        {            b = mull(b, m);        }        n = n>>1;        m = mull(m, m);    }    return b;}int main(){    __int64 n;    while(scanf("%I64d", &n) != EOF)    {        if(n == -1) break;        else if(n == 0)        {           printf("0\n");           continue;        }        __int64 a11, a12, a21, a22;        Matrix m, ans;        m.a11 = 1;        m.a12 = 1;        m.a21 = 1;        m.a22 = 0;        ans = find(m, n);        __int64 result = ans.a12%10000;        printf("%I64d\n", result);    }    return 0;}