• Matrix Power Series


    题目
    其中(A)是一个(n imes n)的矩阵,(S_k = A + A^2 + A^3 + … + A^k),求(S_k)
    按照数论出现和,那么构造一个和数论一起递推的式子(S_k = S_{k - 1} + A^k)
    那么假设(A)不是矩阵

    [left[egin{array}{l} 1 & 1\ 0 & A end{array} ight] imes left[egin{array}{l} S_{k - 1}\ A^k end{array} ight] = left[egin{array}{l} S_k\ A^{k + 1} end{array} ight]]

    用E代替1,用矩阵代替数字,转换一下就是

    [left[egin{array}{l} E & E\ 0 & A end{array} ight] imes left[egin{array}{l} S_{k - 1}\ A^k end{array} ight] = left[egin{array}{l} S_k\ A^{k + 1} end{array} ight] ]

    构建了个分块矩阵,大小是(2n imes 2n)
    加速矩阵(left[egin{array}{l} E & E\ 0 & A end{array} ight]) 初始矩阵(left[egin{array}{l} S_1\ A^2 end{array} ight])

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #define ll long long
    using namespace std;
    const int N = 61;
    int mod, k, n;
    struct Matrix{//矩阵
        int n,m;
        int a[N][N];
        Matrix(int x,int y):n(x),m(y){memset(a,0,sizeof(a));}
        Matrix operator * (const Matrix &b){
            Matrix ans(n,b.m);
            for(int i = 0; i < n; i++){
                for(int j = 0; j < b.m; j++){
                    for(int k = 0; k < m; k++){
                        ans.a[i][j] = (ans.a[i][j] + a[i][k] * b.a[k][j] % mod) % mod;
                    }
                }
            }
            return ans;
        }
    };
    Matrix ksm(Matrix a, ll b){
        Matrix ans(a.n, a.m);
        for(int i = 0; i <= max(a.n, a.m); i++)
            ans.a[i][i] = 1;
    
        while(b){
            if(b & 1)ans = ans * a;
            a = a * a;
            b >>= 1;
        }
        return ans;
    }
    Matrix a(35, 35);
    void solve(){
        Matrix base(k * 2, k * 2);
        for(int i = 0; i < k; i++)
            base.a[i][i] = base.a[i][i + k] = 1;
    
        for(int i = k; i < 2 * k; i++)
            for(int j = k; j < 2 * k; j++)
                base.a[i][j] = a.a[i - k][j - k];
    
        base = ksm(base, n - 1);
        
    
        Matrix ans(k * 2, k);
        for(int i = 0; i < k; i++)
            for(int j = 0; j < k; j++)
                ans.a[i][j] = a.a[i][j];
    
        a = a * a;
        for(int i = k; i < 2 * k; i++)
            for(int j = 0; j < k; j++)
                ans.a[i][j] = a.a[i - k][j];
    
        ans = base * ans;
        for(int i = 0; i < k; i++){
            for(int j = 0; j < k; j++)
                printf("%d ", ans.a[i][j]);
            putchar('
    ');
        }
    }
    int main(){
        scanf("%d%d%d", &k, &n, &mod);
        for(int i = 0; i < k; i++)
            for(int j = 0; j < k; j++)
                scanf("%d", &a.a[i][j]);
        solve();
        return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/Emcikem/p/12891573.html
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