HDU 3483 A Very Simple Problem 矩阵构造

利用二项式定理，可以可以将x^k*k^x展开得x^(k-1)*(1+k-1)^x从而得到递推关系，构造出矩阵直接快速幂就好了。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <climits>
#include <iostream>
#include <string>

using namespace std;
 
#define MP make_pair
#define PB push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int INF = INT_MAX / 3;
const double eps = 1e-8;
const LL LINF = 1e17;
const double DINF = 1e60;
const int maxn = 65;
int N, M, x;

struct Matrix {
    int n, m;
    LL data[maxn][maxn];
    Matrix(int n = 0, int m = 0): n(n), m(m) {
        memset(data, 0, sizeof(data));
    }
    void print() {
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                cout << data[i][j] << " ";
            }
            cout << endl;
        }
    }
};

Matrix operator * (Matrix a, Matrix b) {
    Matrix ret(a.n, b.m);
    for(int i = 1; i <= a.n; i++) {
        for(int j = 1; j <= b.m; j++) {
            for(int k = 1; k <= a.m; k++) {
                ret.data[i][j] += a.data[i][k] * b.data[k][j];
                ret.data[i][j] %= M;
            }
        }
    }
    return ret;
}

Matrix pow(Matrix val, LL p) {
    Matrix ret(val.n, val.m);
    for(int i = 1; i <= val.n; i++) ret.data[i][i] = 1;
    while(p) {
        if(p & 1) ret = ret * val;
        val = val * val;
        p >>= 1;
    }
    return ret;
}

int main() {
    while(cin >> N >> x >> M) {
        if(N < 0) break;
        Matrix A0(x + 2, 1), A(x + 2, x + 2);
        for(int i = 1; i <= x + 1; i++) A0.data[i][1] = x;
        A.data[x + 1][1] = 0; A.data[1][1] = x;
        for(int i = 2; i <= x + 1; i++) {
            for(int j = 1; j <= i; j++) {
                A.data[i][j] = (A.data[i - 1][j] + A.data[i - 1][j - 1]) % M;
            }
        }
        A.data[x + 2][x + 1] = A.data[x + 2][x + 2] = 1;
        A0 = pow(A, N) * A0;
        cout << A0.data[x + 2][1] << endl;
    }
    return 0;
}