uva10689 yet another number Sequence

Problem B
Yet another number Sequence
Input:standard input
Output:standard output
Time Limit:3 seconds

Let's define another number sequence, given by the following function:
F (0) = a
F (1) = b
F (n) = f (n-1) + f (n-2), n>1
When a = 0 and B = 1, this sequence gives the Fibonacci sequence. Changing the values of A and
b, you can get many different sequences. Given the values of a, B, you had to find the last m
Digits of f (n).
Input

The first line gives the number of test cases, which was less than 10001. Each test case consists of
A single line containing the integers a B N m. The values of A and B range in [0,100], value of n
ranges in [0, 1000000000] and value of M ranges in [1, 4].

Output
For each test case, print the last m digits of f (n). However, you should not print any leading
Zero.

Sample input Output for sample input
4
0 1 11 3
0 1 42 4
0 1 22 4
0 1 21 4
89
4296
7711
946

Problem Setter:sadrul Habib Chowdhury
Special Thanks:derek Kisman, Member of Elite problem Setters ' Panel

/* is the fast power Fibonacci number refueling ... Time:2015-4-8 16:14 * * #include <cstdio> #include <cstring> #include <cmath> #include <algorithm
> Using namespace std;
int mod;
    struct matrix{int mat[2][2];
    int n;
        void Init (int _n) {n=_n;
    memset (Mat,0,sizeof (MAT)); } Matrix operator * (const matrix &AMP;B) const{matrix ret; ret.
        Init (n);
                    for (int i=0;i<n;i++) {for (int. j=0;j<n;j++) {for (int k=0;k<n;k++) {
                    RET.MAT[I][J]+=MAT[I][K]*B.MAT[K][J];
                Ret.mat[i][j]%=mod;
    }}} return ret;
}
};
    Matrix Pow_mat (Matrix A,int k) {matrix ret; Ret.
    Init (2);
    Ret.mat[0][0]=1;ret.mat[1][1]=1;

        while (k>0) {if (k&1) ret=ret*a;
        A=a*a;
    k>>=1;
} return ret;
    } int main () {int T;
    int a,b,c,d;
    Matrix F,trans; Trans.
    Init (2); F.init (2); trans.mat[0][0]=0; Trans.mat[0][1]=trans. mat[1][0]=trans.mat[1][1]=1;
    scanf ("%d", &t);
        while (t--) {scanf ("%d%d%d%d", &a,&b,&c,&d);
        Mod=1;while (d--) {mod*=10;}

        F.mat[0][0]=a;f.mat[0][1]=b;
        Matrix Ret=pow_mat (TRANS,C);
            /* for (int i=0;i<2;i++) {for (int j=0;j<2;j++) printf ("%d", ret.mat[i][j]);
        Puts ("");
        }*/F=f*ret;
    printf ("%d\n", f.mat[0][0]);
} return 0;
 }