Poj 3233 matrix power series (matrix power)

DefaultMatrix Power Series Time limit:3000 Ms Memory limit:131072 K Total submissions:15553 Accepted:6658 Description GivenN×NMatrixAAnd a positive integerK, Find the sumS=A+A2 +A3 +... +AK. Input The input contains exactly one test case. The first line of input contains three positive integersN(N≤ 30 ),K(K≤ 109) andM(M<104). Then followNLines each containingNNonnegative integers below 32,768, givingA'S elements in row-Major Order. Output Output the elementsSModuloMIn the same wayAIs given. Sample Input 2 2 40 11 1 Sample output 1 22 3 Source

Question: give an n * n matrix A, and find a + A ^ 2 + A ^ 3 + ...... + What is the result of a ^ K mod m?
Method 1: two decimal points
SK = a + A2 + A3 +... + AK
= (1 + a ^ (K/2) * (a + A ^ 2 + A ^ 3 +... + A ^ (K/2) + {A ^ k}
= (1 + a ^ (K/2) * (S (K/2) + {AK} When k is an even number, no {AK}
That is
K % 2 = 0: s [k] = f [k/2] (1 + a [k/2]);
K % 2 = 1: s [k] = f [k-1] + A [k];
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 32;struct Matrix {    int mat[N][N];    Matrix() {        memset(mat, 0, sizeof(mat));        for(int i = 0; i < N; i++)            mat[i][i] = 1;    }} E;int n, k, mod;Matrix Multi(Matrix a, Matrix b) {    Matrix res;    for(int i = 0; i < n; i++) {        for(int j = 0; j < n; j++) {            res.mat[i][j] = 0;            for(int k = 0; k < n; k++)                res.mat[i][j] = (res.mat[i][j] + a.mat[i][k] * b.mat[k][j]) % mod;        }    }    return res;}Matrix Add(Matrix a, Matrix b) {    Matrix res;    for(int i = 0; i < n; i++)        for(int j = 0; j < n; j++)            res.mat[i][j] = (a.mat[i][j] + b.mat[i][j]) % mod;    return res;}Matrix Pow(Matrix a, int n) {    Matrix res;    while(n) {        if(n&1) res = Multi(res, a);        a = Multi(a, a);        n >>= 1;    }    return res;}Matrix Get_Ans(Matrix a, int k) {    if(k == 1) return a;    if(k&1) return Add(Pow(a, k), Get_Ans(a, k-1));    if(k % 2 == 0) {        Matrix A = Get_Ans(a, k/2);        Matrix B = Pow(a, k/2);        Matrix C = Multi(A, B);        return Add(C, A);    }}int main() {    Matrix A;    while(~scanf("%d%d%d", &n, &k, &mod)) {        for(int i = 0; i < n; i++)            for(int j = 0; j < n; j++) {                scanf("%d", &A.mat[i][j]);                A.mat[i][j] %= mod;            }        Matrix ans = Get_Ans(A, k);        for(int i = 0; i < n; i++) {            for(int j = 0; j < n; j++) {                if(j) printf(" ");                printf("%d", ans.mat[i][j]);            }            printf("\n");        }    }    return 0;}


Method 2:
Construct a new matrix B = | A E |
| 0 E |

Then B ^ (k + 1) = | a ^ (k + 1) a ^ K + A ^ (k-1) + ...... + A ^ 2 + A + 1 |
| 0 E |
Therefore, you can use the Matrix to quickly obtain the power of B ^ (k + 1), and then subtract the unit matrix from the part in the upper left corner to obtain the final required matrix.
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 61;int n, mod, k;struct Matrix {    int mat[N][N];    Matrix() {        memset(mat, 0, sizeof(mat));        for(int i = 0; i < N; i++)            mat[i][i] = 1;    }};Matrix Multi(Matrix a, Matrix b) {    Matrix res;    memset(res.mat, 0, sizeof(res.mat));    for(int i = 0; i < n * 2; i++) {        for(int k = 0; k < n * 2; k++) {            if(a.mat[i][k]) {                for(int j = 0; j < n * 2; j++) {                    res.mat[i][j] += a.mat[i][k] * b.mat[k][j];                    res.mat[i][j] %= mod;                }            }        }    }    return res;}Matrix Pow(Matrix x, int m) {    Matrix res;    while(m) {        if(m&1) res = Multi(res, x);        x = Multi(x, x);        m >>= 1;    }    return res;}int main() {    Matrix A;    while(~scanf("%d%d%d",&n, &k, &mod)) {        memset(A.mat, 0, sizeof(A.mat));        for(int i = 0; i < n; i++) {            for(int j = 0; j < n; j++) {                scanf("%d", &A.mat[i][j]);                A.mat[i][j] %= mod;            }            A.mat[i][i+n] = 1;        }        for(int i = n; i < 2 * n; i++)            A.mat[i][i] = 1;        Matrix ans = Pow(A, k + 1);        for(int i = 0; i < n; i++) {            for(int j = n; j < n * 2; j++) {                if(j > n) printf(" ");                if(j - i == n)                    printf("%d", (ans.mat[i][j] - 1 + mod) % mod);                else                    printf("%d", ans.mat[i][j]);            }            printf("\n");        }    }    return 0;}

Poj 3233 matrix power series (matrix power)