快速矩阵幂POJ3233

矩阵幂运算与模运算

最新推荐文章于 2020-12-11 12:06:47 发布

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给出n,k,m；n为矩阵A的行数和列数。然后输入矩阵A

求A^1+A^2+...A^k 全部mod m的矩阵。

#include<cstdio>

int n,mod, k;

struct Matrix
{
    int m[32][32];
}E, Z;

Matrix Mut(Matrix A, Matrix B)
{
    Matrix ans;
    for (int i = 0; i<n; i++)
        for (int j = 0; j<n; j++)
        {
            ans.m[i][j] = 0;
            for (int k = 0; k<n; k++)
            {
                ans.m[i][j] += ((A.m[i][k])*(B.m[k][j]));
                ans.m[i][j] %= mod;
            }
        }
    return ans;
}
Matrix Add(Matrix A, Matrix B)
{
    Matrix ans;
    for (int i = 0; i<n; i++)
        for (int j = 0; j<n; j++)
            ans.m[i][j] = (A.m[i][j] + B.m[i][j]) % mod;
    return ans;
}
Matrix Pow(Matrix A, int b)
{
    Matrix t = A, ans = E;
    while (b)
    {
        if (b % 2) ans = Mut(ans, t);
        b /= 2;
        t = Mut(t, t);
    }
    return ans;
}

Matrix solve(Matrix A, int b)
{
    if (b == 0) return E;
    else if (b == 1) return A;
    else if (b == 2) return Add(A, Mut(A, A));
    else if (b % 2 == 1) return Add(Pow(A, b), solve(A, b - 1));
    else if (b % 2 == 0)
    {
        Matrix ans = solve(A, b / 2);
        Matrix C = Pow(A, b / 2);
        Matrix B = Mut(ans, C);
        return Add(B, ans);
    }
}

int main()
{
    // freopen("in.txt", "r", stdin);

	Matrix A;
   while(scanf("%d %d %d",&n,&k,&mod)!=EOF)
   { 
        for (int i = 0; i<n; i++)
            for (int j = 0; j<n; j++)
                E.m[i][j] = (i == j);

		for(int i=0;i<n;i++)
			for(int j=0;j<n;j++)
				scanf("%d",&A.m[i][j]);

        Matrix ans = solve(A, k);

		for(int i=0;i<n;i++)
		{
			for(int j=0;j<n;j++)
			{
				printf("%d",ans.m[i][j]);
				if(j==n-1)
					printf("\n");
				else 
					printf(" ");
			}
		}
    }
    return 0;
}