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  • HDU 4965 Fast Matrix Calculation(矩阵高速幂)

    HDU 4965 Fast Matrix Calculation

    题目链接

    矩阵相乘为AxBxAxB...乘nn次。能够变成Ax(BxAxBxA...)xB,中间乘n n - 1次,这样中间的矩阵一个仅仅有6x6。就能够用矩阵高速幂搞了

    代码:

    #include <cstdio>
    #include <cstring>
    
    const int N = 1005;
    const int M = 10;
    int n, m;
    
    int A[N][M], B[M][N], C[M][M], CC[N][N];
    int ans[M][M];
    
    void tra() {
        memset(CC, 0, sizeof(CC));
        for (int i = 0; i < m; i++) {
        for (int j = 0; j < m; j++) {
            CC[i][j] = 0;
            for (int k = 0; k < m; k++) {
            CC[i][j] = (CC[i][j] + C[i][k] * C[k][j]) % 6;
            }
        }
        }
        for (int i = 0; i < m; i++)
        for (int j = 0; j < m; j++)
            C[i][j] = CC[i][j];
    }
    
    void mul() {
        for (int i = 0; i < m; i++) {
        for (int j = 0; j < m; j++) {
            CC[i][j] = 0;
            for (int k = 0; k < m; k++) {
            CC[i][j] = (CC[i][j] + ans[i][k] * C[k][j]) % 6;
            }
        }
        }
        for (int i = 0; i < m; i++)
        for (int j = 0; j < m; j++)
            ans[i][j] = CC[i][j];
    }
    
    void pow_mod(int k) {
        memset(ans, 0, sizeof(ans));
        for (int i = 0; i < m; i++)
        ans[i][i] = 1;
        while (k) {
        if (k&1) mul();
        tra();
        k >>= 1;
        }
    }
    
    void init() {
        for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            scanf("%d", &A[i][j]);
        for (int i = 0; i < m; i++)
        for (int j = 0; j < n; j++)
            scanf("%d", &B[i][j]);
    }
    
    int solve() {
        for (int i = 0; i < m; i++) {
        for (int j = 0; j < m; j++) {
            C[i][j] = 0;
            for (int k = 0; k < n; k++) {
            C[i][j] = (C[i][j] + B[i][k] * A[k][j]) % 6;
            }
        }
        }
    
        pow_mod(n * n - 1);
    
        for (int i = 0; i < m; i++) {
        for (int j = 0; j < m; j++) {
            C[i][j] = ans[i][j];
        }
        }
    
        for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            CC[i][j] = 0;
            for (int k = 0; k < m; k++) {
            CC[i][j] = (CC[i][j] + A[i][k] * C[k][j]) % 6;
            }
        }
        }
        for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            A[i][j] = CC[i][j];
        int ans = 0;
        for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            int sum = 0;
            for (int k = 0; k < m; k++) {
            sum = (sum + A[i][k] * B[k][j]) % 6;
            }
            ans += sum;
        }
        }
        return ans;
    }
    
    int main() {
        while (~scanf("%d%d", &n, &m) && n || m) {
        init();
        printf("%d
    ", solve());
        }
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/gcczhongduan/p/5193591.html
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