POJ3070

1.题目描述：

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 14467 Accepted: 10217

Description

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

.

Source

Stanford Local 2006

2.题目描述：

题目大意就是求出斐波那契数列第k项的后四位，如果有前导0就不输出0，但最后一个0必须输出

刚开始的时候就想到了一个关于斐波那契数列的数列的一个公式，准备用那个公式求的，但题目已经提示了用矩阵，就没用那个公式

有兴趣的可以看看这个公式，试试能不能那个解这题

这一题就要注意到矩阵

的右上角或左下角对应的值正好是第n项的值

故可以通过快速幂来求得斐波那契数列的哦n项，由于数据较大，还要记得要不断的对10000取模

3.解题思路：

题目把矩阵已经构造好了，直接套矩阵快速幂模板就行

4.AC代码：

#include <cstdio>#include <iostream>using namespace std;const int MOD = 10000;struct matrix{    int m[2][2];}ans, base;matrix multi(matrix a, matrix b){    matrix tmp;    for(int i = 0; i < 2; ++i)    {        for(int j = 0; j < 2; ++j)        {            tmp.m[i][j] = 0;            for(int k = 0; k < 2; ++k)                tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;        }    }    return tmp;}int fast_mod(int n)  // 求矩阵 base 的  n 次幂 {    base.m[0][0] = base.m[0][1] = base.m[1][0] = 1;    base.m[1][1] = 0;    ans.m[0][0] = ans.m[1][1] = 1;  // ans 初始化为单位矩阵     ans.m[0][1] = ans.m[1][0] = 0;    while(n)    {        if(n & 1)  //实现 ans *= t; 其中要先把 ans赋值给 tmp，然后用 ans = tmp * t         {            ans = multi(ans, base);        }        base = multi(base, base);        n >>= 1;    }    return ans.m[0][1];}int main(){    int n;    while(scanf("%d", &n) && n != -1)    {           printf("%d\n", fast_mod(n));    }    return 0;}