POJ3070Fibonacci

易水人去，明月如霜。

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:

.

#include <cstdio>#include <iostream>using namespace std;const int MOD = 10000;int fast_mod(int n){    int t[2][2] = {1, 1, 1, 0};    int ans[2][2] = {1, 0, 0, 1};    int tmp[2][2];    while(n)    {        if(n & 1)        {            for(int i = 0; i < 2; ++i)                for(int j = 0; j < 2; ++j)                    tmp[i][j] = ans[i][j];            ans[0][0] = ans[1][1] = ans[0][1] = ans[1][0] = 0;            for(int i = 0; i < 2; ++i)            {                for(int j = 0; j < 2; ++j)                {                    for(int k = 0; k < 2; ++k)                        ans[i][j] = (ans[i][j] + tmp[i][k] * t[k][j]) % MOD;                }            }        }        for(int i = 0; i < 2; ++i)            for(int j = 0; j < 2; ++j)                tmp[i][j] = t[i][j];        t[0][0] = t[1][1] = 0;        t[0][1] = t[1][0] = 0;        for(int i = 0; i < 2; ++i)        {            for(int j = 0; j < 2; ++j)            {                for(int k = 0; k < 2; ++k)                    t[i][j] = (t[i][j] + tmp[i][k] * tmp[k][j]) % MOD;            }        }        n >>= 1;    }    return ans[0][1];}int main(){    int n;    while(scanf("%d", &n) && n != -1)    {        printf("%d\n", fast_mod(n));    }    return 0;}