UVa 10229 Modular Fibonacci: Matrix fast Power seeking Fibonacci

10229-modular Fibonacci

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=115&page=show_ problem&problem=1170

The Fibonacci numbers (0, 1, 1, 2, 3, 5, 8, ...) are defined by the recurrence:

F0 = 0 F1 = 1 Fi = Fi-1 + Fi-2 for i>1

Write a program which calculates Mn = Fn mod 2m for given pair of N and M. 0< =n< and =2147483647 20. Note This a mod B gives the remainder when a was divided by B.

Input and Output

Input consists of several lines specifying a pair of N and M. Output should Mn, one per line.

Sample Input

7
11 6

Sample Output

25

Ideas:

As shown above, use the matrix fast power to fix.

Note: (1<<19) ^2 will be super int, so use long long

Complete code:

/*0.015s*/
     
#include<cstdio>  
#include<cstring>  
typedef long long ll;  
     
ll mod;  
     
struct mat  
{  
    ll v[2][2];  
    mat()  
    {  
        memset(v, 0, sizeof(v));  
    }  
} m1;  
     
mat operator * (mat a, mat b)  
{  
    mat c;  
    for (int i = 0; i < 2; i++)  
        for (int j = 0; j < 2; j++)  
            for (int k = 0; k < 2; k++)  
                c.v[i][j] = (c.v[i][j] + a.v[i][k] * b.v[k][j]) % mod;  
    return c;  
}  
     
inline mat quickpow(mat a, int k)  
{  
    mat b;  
    b.v[0][0] = b.v[1][1] = 1;  
    while (k)  
    {  
        if (k & 1)  
            b = b * a;  
        k >>= 1;  
        a = a * a;  
    }  
    return b;  
}  
     
int main()  
{  
    m1.v[0][0] = m1.v[0][1] = m1.v[1][0] = 1, m1.v[1][1] = 0;  
    int n, m;  
    while (~scanf("%d%d", &n, &m))  
    {  
        if (n == 0 || m == 0) puts("0");  
        else
        {  
            mod = 1 << m;  
            printf("%d\n", quickpow(m1, n - 1).v[0][0]);  
        }  
    }  
    return 0;  
}

 

	
  


 

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