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

 

code

//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;
}

 

转载于:https://www.cnblogs.com/superbin/archive/2010/05/14/1735853.html