nyoj 148

链接：

http://acm.nyist.net/JudgeOnline/problem.php?pid=148

fibonacci数列(二)

时间限制：1000 ms  |  内存限制：65535 KB
难度：3

描述

    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.

    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:

    .

输入
    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.
输出
    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).
样例输入

    0
    9
    1000000000
    -1

样例输出

    0
    34

    6875

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

const int mod=10000;  
long x,y,n;  

struct mat  
{  
    int a[2][2];
};  
mat mat_mul(const mat &x,const mat &y)
{  
    static mat res;  
    memset(res.a,0,sizeof(res.a));  
    for(int i=0; i<2; i++)  
        for(int j=0; j<2; j++)  
            for(int k=0; k<2; k++)
                if(x.a[i][k] && y.a[k][j])    
                    res.a[i][j]=(res.a[i][j]+x.a[i][k]*y.a[k][j])%mod;
    return res;  
}
void mat_fast_pow(long n)  
{  
    static int i;
    static mat c,res;  
    c.a[0][0]=c.a[0][1]=c.a[1][0]=1;  
    c.a[1][1]=0;  
    memset(res.a,0,sizeof(res.a));  
    for(i = 0;i < 2;++i)  
        res.a[i][i]=1;  

    while(n)  
    {  
        if(n&1)
            res=mat_mul(res,c);
        c=mat_mul(c,c);  

        n=n>>1;  
    }  
    int yy=res.a[0][0],xx=res.a[0][1];
    printf("%d\n", yy);  
}  
int main()  
{  
    while(~scanf("%I64d", &n))  
    {  
        if(n > 0)
            mat_fast_pow(n-1);  
        else if(n == 0)
            printf("0\n");
        else
            break;
    }  
    return 0;  
}