Matrix 10 point "two" Poj 1575 Tr A poj 3233 Matrix Power Series

POJ 1575 Tr A

Topics Link: http://acm.hdu.edu.cn/showproblem.php?pid=1575

The main idea: A is a square, then tr A is a sign of a (that is, the a^k of the major diagonal), now requires TR (%9973).

The first line of data is a T, which indicates that there is a T group of data.

The first row of each group of data has n (2 <= n <= 10) and K (2 <= K < 10^9) two data. Next there are n rows, each row has n data, and the range of each data is [0,9]. Represents the content of square a.

A naked problem with a matrix high-speed power.

Exercises

#include <iostream> #include <stdio.h> #include <cstring> #define Mod 9973using namespace Std;const int MAX = 11;struct matrix{int v[max][max];}; int N, k, M;    Matrix Mtmul (Matrix A, matrix B)//Seeking matrices A * b{int i, j, K;    Matrix C;            for (i = 0; i < n; i + +) for (j = 0; J < N; j + +) {C.v[i][j] = 0;        for (k = 0; k < n; k + +) C.v[i][j] = (a.v[i][k] * B.v[k][j] + c.v[i][j])% Mod; } return C;}        Matrix Mtpow (Matrix A, int k)//seeking matrix A ^ k{if (k = = 0) {memset (a.v, 0, sizeof (A.V));        for (int i = 0; i < n; i + +) a.v[i][i] = 1;    return A;    } if (k = = 1) return A;    Matrix C = Mtpow (A, K/2);    if (k% 2 = = 0) return Mtmul (c, c); else return Mtmul (Mtmul (c, c), A);    int Solv (Matrix A) {int ans=0;    for (int i=0;i<n;i++) Ans+=a.v[i][i]%mod; return ans;} void out (Matrix A) {for (int. i=0;i<n;i++) {for (int j=0;j< n;j++) printf ("%d", a.v[i][j]);    cout<<endl;    }}int Main () {int T;    scanf ("%d", &t);        while (t--) {scanf ("%d%d", &n,&k);        Matrix A;        for (int i=0;i<n;i++) for (int j=0;j<n;j++) scanf ("%d", &a.v[i][j]);        Matrix ans;        Ans=mtpow (A,K);        Out (ans);    Cout<<solv (ANS)%mod<<endl; }}

The POJ 3233 Matrix power series has the same matrix high-speed power-up problem, which is mainly used for s (6) = (1 + a^3) * S (3) and S (7) = A + (A + a^4) * S (3), and these two last-time recursion will save a lot of time

#include <iostream> #include <cstring> #include <stdio.h>using namespace std;const int MAX = 32;struct matrix{int v[max][max];}; int N, k, M;    Matrix Mtadd (Matrix A, matrix B)//seeking matrices A + b{int i, J;    Matrix C;    for (i = 0; i < n; i + +) for (j = 0; J < N; j + +) c.v[i][j]= (A.v[i][j]+b.v[i][j])% M; return C;}    Matrix Mtmul (Matrix A, matrix B)//Seeking matrices A * b{int i, j, K;    Matrix C;            for (i = 0; i < n; i + +) for (j = 0; J < N; j + +) {C.v[i][j] = 0;        for (k = 0; k < n; k + +) C.v[i][j] = (a.v[i][k] * B.v[k][j] + c.v[i][j])% M; } return C;}        Matrix Mtpow (Matrix A, int k)//seeking matrix A ^ k{if (k = = 0) {memset (a.v, 0, sizeof (A.V));        for (int i = 0; i < n; i + +) a.v[i][i] = 1;    return A;    } if (k = = 1) return A;    Matrix C = Mtpow (A, K/2);    if (k% 2 = = 0) return Mtmul (c, c); else return Mtmul (Mtmul (C, c), A);    Matrix mtcal (Matrix A, int k)//Ask S (k) = A + A2 + A3 + ... + ak{if (k = = 1) return A;    Matrix B = Mtpow (A, (k+1)/2);    Matrix C = mtcal (A, K/2);   if (k% 2 = = 0) return Mtmul (Mtadd (Mtpow (A, 0), B), C); such as s (6) = (1 + a^3) * S (3).
else return Mtadd (A, Mtmul (Mtadd (A, B), C));        such as s (7) = A + (A + a^4) * S (3)}void out (Matrix A) {for (int i=0;i<n;i++) {for (int j=0;j<n-1;j++)        cout<<a.v[i][j]<< "";    cout<<a.v[i][n-1]<<endl;    }}int Main () {Matrix A;    scanf ("%d%d%d", &n,&k,&m);    for (int i=0;i<n;i++) for (int j=0;j<n;j++) scanf ("%d", &a.v[i][j]);    Matrix c=mtcal (a,k); Out (C);}

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Matrix 10 point "two" Poj 1575 Tr A poj 3233 Matrix Power Series