UVA 11149 Power of Matrix

Posted 2020-06-17

矩阵快速幂。

读入A矩阵之后，马上对A矩阵每一个元素%10，否则会WA.....

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;

int MOD=10;
int n,m;

struct Matrix
{
    int A[90][90];
    int R, C;
    Matrix operator*(Matrix b);
};

Matrix A, X, Y, Z;

int mod(int a, int b)
{
    if (a >= 0) return a%b;
    if (abs(a) % b == 0) return 0;
    return (a + b*(abs(a) / b + 1));
}

Matrix Matrix::operator*(Matrix b)
{
    Matrix c;
    memset(c.A, 0, sizeof(c.A));
    int i, j, k;
    for (i = 1; i <= R; i++)
        for (j = 1; j <= C; j++)
            for (k = 1; k <= C; k++)
                c.A[i][j] = mod((c.A[i][j] + mod(A[i][k] * b.A[k][j], MOD)), MOD);
    c.R=R; c.C=b.C;
    return c;
}

void read()
{
    A.R=A.C=n;
    for(int i=1;i<=n;i++) 
        for(int j=1;j<=n;j++) {
            scanf("%d",&A.A[i][j]);
            A.A[i][j]=A.A[i][j]%MOD;
    }
}

void init()
{
   m=m-1;
   memset(Y.A,0,sizeof Y.A);
   memset(Z.A,0,sizeof Z.A);
   memset(X.A,0,sizeof X.A);

   Y.R=2*n; Y.C=2*n;
   for(int i=1;i<=2*n;i++) Y.A[i][i]=1;

   Z.R=n; Z.C=2*n;
   for(int i=1;i<=n;i++) Z.A[i][i]=1;
   for(int i=1;i<=n;i++) for(int j=n+1;j<=2*n;j++) Z.A[i][j]=A.A[i][j-n];

   X.R=2*n; X.C=2*n;
   for(int i=1;i<=n;i++) X.A[i][i]=1;
   for(int i=1;i<=n;i++) for(int j=n+1;j<=2*n;j++) X.A[i][j]=A.A[i][j-n];
   for(int i=n+1;i<=2*n;i++) for(int j=n+1;j<=2*n;j++) X.A[i][j]=A.A[i-n][j-n];
}

void work()
{
    while (m)
    {
        if (m % 2 == 1) Y = Y*X;
        m = m >> 1;
        X = X*X;
    }
    Z = Z*Y;

    for(int i=1;i<=n;i++)
    {
        for(int j=n+1;j<=2*n;j++)
        {
            printf("%d",Z.A[i][j]%MOD);
            if(j<2*n) printf(" ");
            else printf("\n");
        }
    }
    printf("\n");
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(!n) continue;
        read();
        init();
        work();
    }
    return 0;
}