poj 3070 Fibonacci 【矩阵快速幂】

Fibonacci
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 10072 Accepted: 7191
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 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 Fn.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).

Sample Input

0
9
999999999
1000000000
-1
Sample Output

0
34
626
6875
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

做的第一道矩阵，先从别人那里盗个模板。。
矩阵快速幂用来计算矩阵的n次方的。将时间复杂度降到log（n），原理和快速幂类似，二分的思想（想不到当年学的线性代数用到了，orz）；主要在于构造矩阵；

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <ctype.h>
#include <time.h>
#include <queue>

using namespace std;

const int MOD = 10000;

struct node
{
    int m[2][2];
}ans,base;

int n;

node multi(node a,node b)
{
    node 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] +=(a.m[i][k] * b.m[k][j]);
            tmp.m[i][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()
{
    while (scanf("%d",&n)!=EOF)
    {
        if (n == -1)
            break;
        int ans = fast_mod(n);
        printf("%d\n",ans);
    }
    return 0;
}