POJ 3233 Matrix Power Series(矩阵快速幂+二分)

Time Limit: 3000MS		Memory Limit: 131072K
Total Submissions: 16986		Accepted: 7233

Description

Given a n × n matrix A and a positive integer k, find the sum S = A + A² + A³ + … + A^k.

Input

The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 10⁹) and m (m < 10⁴). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.

Output

Output the elements of S modulo m in the same way as A is given.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3


解题思路：
代码：

# include<iostream>
# include<cstdio>
# include<cstring>
using namespace std;
const int MAX = 32;

struct Matrix
{
    int v[MAX][MAX];
};
int n, k, M;

Matrix mtAdd(Matrix A, Matrix B)        // 求矩阵 A + B
{
    int i, j;
    Matrix C;
    for(i = 0; i < n; i ++)
        for(j = 0; j < n; j ++)
        {
            C.v[i][j] = (A.v[i][j] + B.v[i][j]) % M;
        }
    return C;
}

Matrix mtMul(Matrix A, Matrix B)        // 求矩阵 A * B
{
    int i, j, k;
    Matrix C;
    for(i = 0; i < n; i ++)
        for(j = 0; j < n; j ++)
        {
            C.v[i][j] = 0;
            for(k = 0; k < n; k ++)
            {
                C.v[i][j] = (A.v[i][k] * B.v[k][j] + C.v[i][j]) % M;
            }
        }
    return C;
}

Matrix mtPow(Matrix A, int k)           // 求矩阵 A ^ k
{
    if(k == 0) {
        memset(A.v, 0, sizeof(A.v));
        for(int i = 0; i < n; i ++)
        {
            A.v[i][i] = 1;
        }
        return A;
    }
    if(k == 1) return A;
    Matrix C = mtPow(A, k / 2);
    if(k % 2 == 0) {
        return mtMul(C, C);
    } else {
        return mtMul(mtMul(C, C), A);
    }
}

Matrix mtCal(Matrix A, int k)           // 求S (k) = A + A2 + A3 + … + Ak
{
    if(k == 1) return A;
    Matrix B = mtPow(A, (k+1) / 2);
    Matrix C = mtCal(A, k / 2);
    if(k % 2 == 0) {
        return mtMul(mtAdd(mtPow(A, 0), B), C);   // 如S(6) = (1 + A^3) * S(3)。
    } else {
        return mtAdd(A, mtMul(mtAdd(A, B), C));   // 如S(7) = A + (A + A^4) * S(3)
    }
}

int main(void)
{
    int i, j;
    Matrix A;
    cin >> n >> k >> M;
    for(i = 0; i < n; i ++)
        for(j = 0; j < n; j ++)
        {
           scanf("%d",&A.v[i][j]);
        }
    A = mtCal(A, k);
    for(i = 0; i < n; i ++)
    {
        for(j = 0; j < n; j ++)
        {
            cout << A.v[i][j] << ' ';
        }
        cout << endl;
    }
    return 0;
}