hdu-4549 (Matrix fast Power + Euler's theorem)

Problem Description:

M Fibonacci Sequence F[n] is an integer sequence, which is defined as follows:

F[0] = a
F[1] = b
F[n] = f[n-1] * F[n-2] (n > 1)

Now give a, B, N, can you find the value of F[n]?

Input

Input contains multiple sets of test data;
One row per group of data, containing 3 integers a, b, n (0 <= A, b, n <= 10^9)

Output

For each set of test data output an integer f[n], because f[n] may be large, you only need to output F[N] to 1000000007 modulo value, each set of data output one row.

Sample Input

0 1 0
6 10 2

Sample Output

0
60

Topic Test Instructions: Topic give us a n let's find out a new Fibonacci sequence of f (n)% 1e9+7.

Title Analysis: Although this topic is different from the original Fibonacci sequence, but the essence is the same, for a, b their index conforms to the Fibonacci sequence of rules f (n) =f (n-1) +f (n-2), just need to note that their F (0) and F (1) value is not the same, We use matrices to quickly run A and b exponentially, and the rest is to calculate a^ (x1)%c and b^ (x2)%c.

This place I am a lot of fat, find this place is the difficulty of this problem.

For A^b%c, we

The first thing I thought was:

For the a^x% mod formula, we have the following expansion:

A^x% mod=a^ (X%phi[mod]+phi[mod])%mod (when and only when X>=phi[mod] (in my previous blog, I used the formula)

But this place is not good, because the MoD is very large and has no effect.

Here we will use Euler's theorem to solve: in number theory, Euler's theorem, (also called Fermat-Euler theorem) is a property about congruence. Euler's theorem shows that if n,a is a positive integer and N,a coprime, then:

For A^b%c, we make x*phi[c]+y=b equal to a^ (x*phi[c]+y)%c
a^ (x*phi[c]+y)%c=a^ (x*phi[c])%c * a^ (y)%c = (a^ (phi[c])%c) ^x *a^ (y)%c found no, the preceding equation is that the Euler theorem expression equals 1
Then the primitive is equal to a^ (y)%c because C is the prime number then phi[c]=c-1, then y=b% (Phi[c]) =b% (C-1); namely: A^b%c =a^ (B% (C-1))%c code is as follows:

#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #define LL Long Long

using namespace Std;
const int mod=1e9+7;

int a,b,n;
    struct matrix//build matrix {ll f[3][3];
        Matrix operator * (const matrix &a) Const {Matrix res;
                for (int i=1;i<=2;i++) {for (int j=1;j<=2;j++) {res.f[i][j]=0;
                for (int k=1;k<=2;k++) res.f[i][j]= (res.f[i][j]+ (*this). F[i][k]*a.f[k][j]);
            res.f[i][j]=res.f[i][j]% (mod-1);
    }} return res;

}}ppow;
    void init ()//constant matrix {ppow.f[1][1]=ppow.f[1][2]=1;
Ppow.f[2][1]=1;
    } Matrix Fast_pow1 (Matrix Base,int k)//fast power matrix {matrix ans=base;
        while (k) {if (k&1) ans=ans*base;
        Base=base*base;
    k>>=1;
} return ans;
    } ll Fast_pow2 (ll base,ll k)//fast power number {LL ans=1;
    Base=base%mod;
          while (k) {if (k&1)  Ans=ans*base%mod;
        Base=base*base%mod;
    k>>=1;
} return ans;
    } int main () {init ();
        while (scanf ("%d%d%d", &a,&b,&n)!=eof) {if (n==0) {printf ("%d\n", a%mod); continue;}
        if (n==1) {printf ("%d\n", b%mod); continue;}
        n-=2;
        ll Sum=1;
        struct matrix cur;
        Cur=ppow;
        Cur=fast_pow1 (Cur,n);
        Sum=fast_pow2 (a,cur.f[1][2])%mod;//a and B are not the same as the index of the storage note sum=sum*fast_pow2 (b,cur.f[1][1])%mod;
    printf ("%lld\n", sum);
} return 0;
 }