HDU 5015 233 Matrix 矩阵快速幂

可以构造如下矩阵

A：

1 0 0 0 0 0 0 0 0 0 0 0
1 10 0 0 0 0 0 0 0 0 0 0
1 10 1 0 0 0 0 0 0 0 0 0
1 10 1 1 0 0 0 0 0 0 0 0
1 10 1 1 1 0 0 0 0 0 0 0
1 10 1 1 1 1 0 0 0 0 0 0
1 10 1 1 1 1 1 0 0 0 0 0
1 10 1 1 1 1 1 1 0 0 0 0
1 10 1 1 1 1 1 1 1 0 0 0
1 10 1 1 1 1 1 1 1 1 0 0
1 10 1 1 1 1 1 1 1 1 1 0
1 10 1 1 1 1 1 1 1 1 1 1

A0:

3

233

a0+233

a0+a1+233

a0+a1+a2+233

....

这样就有第m列为A^(m-1)*A0

#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 LL mod = 1e7 + 7;

const int maxn = 15;

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] %= mod;
            }
        }
    }
    return ret;
}

Matrix pow(Matrix mat, LL p) {
    Matrix ret(mat.n, mat.m);
    if(p == 0) {
        for(int i = 1; i <= mat.n; i++) ret.data[i][i] = 1;
        return ret;
    }
    if(p == 1) return mat;
    ret = pow(mat * mat, p / 2);
    if(p & 1) ret = ret * mat;
    return ret;
}

int main() {
    Matrix A(12, 12); A.data[1][1] = 1;
    for(int i = 2; i <= 12; i++) {
        A.data[i][1] = 1; A.data[i][2] = 10;
    }
    for(int i = 3; i <= 12; i++) {
        for(int j = 1, k = 3; j < i - 1; j++, k++) {
            A.data[i][k] = 1;
        }
    }
    A.print();
    LL n, m;
    while(cin >> n >> m) {
        LL a[11] = {0};
        for(int i = 1; i <= n; i++) cin >> a[i];
        Matrix A0(12, 1);
        A0.data[1][1] = 3; A0.data[2][1] = 233;
        for(int i = 3, j = 1; i <= 12; i++, j++) {
            A0.data[i][1] = a[j] + A0.data[i - 1][1];
        }
        Matrix ans = pow(A, m - 1) * A0;
        cout << ans.data[n + 2][1] << endl;
    }
    return 0;
}