[POJ3233] Matrix Power Series（矩阵快速幂）

传送门

k <= 10⁹ 暴力肯定超时

根据矩阵性质，可以发现

S(4) = A¹ + A² + A² * (A¹ + A²)

S(5) = A¹ + A² + A² * (A¹ + A²) + A⁵

所以可以递归二分求解，分别判断 A^k，k 为奇数和偶数的情况

——代码

#include <cstdio>
#include <cstring>

struct Matrix
{
    int a[30][30];
    Matrix()
    {
        memset(a, 0, sizeof(a));
    }
};

int n, k, p;
Matrix I, P;

inline Matrix operator * (const Matrix x, const Matrix y)
{
    Matrix ans;
    int i, j, k;
    for(i = 0; i < n; i++)
        for(j = 0; j < n; j++)
            for(k = 0; k < n; k++)
                ans.a[i][j] = (ans.a[i][j] + x.a[i][k] * y.a[k][j]) % p;
    return ans;
}

inline Matrix operator + (const Matrix x, const Matrix y)
{
    Matrix ans;
    int i, j;
    for(i = 0; i < n; i++)
        for(j = 0; j < n; j++)
            ans.a[i][j] = (ans.a[i][j] + x.a[i][j] + y.a[i][j]) % p;
    return ans;
}

inline Matrix operator ^ (Matrix x, int y)
{
    Matrix ans = I;
    while(y)
    {
        if(y & 1) ans = ans * x;
        x = x * x;
        y >>= 1;
    }
    return ans;
}

inline Matrix work(Matrix x, int y)
{
    if(!y) return P;
    Matrix ans = work(x, y >> 1);
    ans = ans + (x ^ (y >> 1)) * ans;
    if(y & 1) ans = ans + (x ^ y);
    return ans; 
}

int main()
{
    int i, j;
    Matrix ans;
    for(i = 0; i < 30; i++) I.a[i][i] = 1;
    while(~scanf("%d %d %d", &n, &k, &p))
    {
        for(i = 0; i < n; i++)
            for(j = 0; j < n; j++)
                scanf("%d", &ans.a[i][j]);
        ans = work(ans, k);
        for(i = 0; i < n; i++)
        {
                for(j = 0; j < n; j++) printf("%d ", ans.a[i][j]);
                puts("");
           }
    }
    return 0;
}