UVa 10229 Modular Fibonacci

背景：斐波那契数列，求Fib[n] mod 2^m

思想：分治法的应用，构造矩阵

 

//12ms  2010-05-14 22:00:33
//Type: 二分法求幂，Fibonacci (矩阵乘法加速)
#include <stdio.h>
#include <string.h>

int main()
{
    int n, m;
    int bi[21];
    long long f1, f0, tf1, tf0;
    bi[1] = 2;
    for (int i=2; i<=20; i++) 
        bi[i] = bi[i-1]<<1;
    while (scanf("%d%d", &n, &m) != EOF) {
        if (!n || !m) { printf("0\n"); continue; }
        if (n == 1) { printf("1\n"); continue; }
        int M = bi[m];
        int p = n-1;
        long long a1, a2, a3, a4, b1, b2, b3, b4;
        a1 = a2 = a3 = 1;
        a4 = 0;
        f1 = 1;
        f0 = 0;
        while (p>0) {
            if (p&1) {
                tf1 = (f1*a1+f0*a2)%M;
                tf0 = (f1*a3+f0*a4)%M;
                f1 = tf1;
                f0 = tf0;
            }
            b1 = (a1*a1+a2*a3)%M;
            b2 = (a1*a2+a2*a4)%M;
            b3 = (a1*a3+a3*a4)%M;
            b4 = (a2*a3+a4*a4)%M;
            a1 = b1;
            a2 = b2;
            a3 = b3;
            a4 = b4;
            p /= 2;
        }
        printf("%lld\n", f1);
    }
    return 0;
}