Fast power of Matrix: NetEase 2017 intern programming problem Magic bracelet __ Programming

Describe:
Small easy to have a magic bracelet with n numbers (forming a ring), when the Magic bracelet uses magic every time a strange change occurs: Each number will become itself with the back of a number of the and (the last number after a number is the first), Once a position has a number greater than or equal to 100, the 100 modulo (for example, a position becomes 103, will automatically change to 3). Now give the Magic bracelet composition, please calculate the Magic bracelet after the use of K magic the state of the ring.
Input Description:
The input data consists of two lines:
The first behavior is two integers n (2≤n≤50) and K (1≤k≤2000000000), separated by a space
The second action magic Bracelet initial n number, separated by a space. The range is 0 to 99.
Output Description:
The output Magic Bracelet uses the state after K times, separated by a space, with no spaces at the end of the line.
Input Example:
3 2
1 2 3
Output Example:
8 9 7

Analysis:
Because the evaluation of the problem is very regular, the number into a ring, and the current number is equal to the current number plus a number behind, naturally can be associated with the structure of the matrix, and see the number of operations so large, must either find the law, or fast power, and then I hit the table did not find any rules, so the matrix fast power.

#include <bits/stdc++.h> using namespace std; vector<vector<int> > Mul (vector<vector<int> >a,vector<vector<int> >b) {int M=A.S
    Ize (), N=a[0].size (), k=b[0].size ();
    vector<vector<int> > Res (m,vector<int> (k,0)); for (int i=0;i<m;i++) for (int j=0;j<k;j++) for (int f=0;f<n;f++) res[i][j] = (res[i][j]+a[i
    ][F]*B[F][J])%100;
return res;
    } vector<vector<int> > Matrix_pow (vector<vector<int> > Matrix,int k) {int n=matrix.size ();
    vector<vector<int> > mm (n,vector<int> (n,0));

    for (int i=0;i<n;i++) mm[i][i]=1;
        while (k) {if (k&1) mm = Mul (MATRIX,MM);
        Matrix = Mul (Matrix,matrix);
    k>>=1;
} return mm;
    } int main () {int n,k;
    scanf ("%d%d", &n,&k);
    vector<vector<int> > num;
    Vector<int> T;
    int tmp; for (int i=0;i<n;i++) {scanf ("%D ", &tmp);
    T.push_back (TMP);
    } num.push_back (t);
    vector<vector<int> > Matrix (n,vector<int> (n,0));
        for (int i=0;i<n;i++) {matrix[i][i]=1;
    matrix[(i+1)%n][i]=1;
    } vector<vector<int> > mm = Matrix_pow (matrix,k);
    vector<vector<int> > ans = mul (num,mm);
    printf ("%d", ans[0][0]);
    for (int i=1;i<ans[0].size (); i++) printf ("%d", ans[0][i]);

    Puts ("");
return 0;
 }