HDU - 5015 233 Matrix (矩阵构造)

In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333... (it means a_0,1 = 233,a_0,2 = 2333,a_0,3 = 23333...) Besides, in 233 matrix, we got a_i,j = a_i-1,j +a_i,j-1( i,j ≠ 0). Now you have known a_1,0,a_2,0,...,a_n,0, could you tell me a_n,m in the 233 matrix?

There are multiple test cases. Please process till EOF.

For each case, the first line contains two postive integers n,m(n ≤ 10,m ≤ 10⁹). The second line contains n integers, a_1,0,a_2,0,...,a_n,0(0 ≤ a_i,0 < 2³¹).

For each case, output a_n,m mod 10000007.

1 1
1
2 2
0 0
3 7
23 47 16

234
2799
72937

Hint

 

2014 ACM/ICPC Asia Regional Xi‘an Online
题意：求最后一位的值
思路：将第i行之后的矩阵都往后移动i位构造一个矩阵：
                                                                                                  10 0 0 0 0 ... 1
                                                                                                  1 1 0 0 0  ..... 0
                                                                                                  0 1 1 0 ...       0
                                                                                                  . . .. . .. . .. .. .. . .
                                                                                                 0 0 0 0 0 ....    1,
但是还需要推出一个B维度为n+2 * 1的初始矩阵，就是根据移动的推出初始的一列

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef __int64 ll;
using namespace std;
const int maxn = 20;
const int mod = 10000007;

struct Matrix{
	int n;
	ll v[maxn][maxn];	

	Matrix(int _n = maxn) {
		n = _n;
	}

	void init(ll _v = 0) {
		memset(v, 0, sizeof(v));
		if (_v)
			for (int i = 0; i < n; i++)
				v[i][i] = _v;
	}

	void output() {
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++)
				printf("%I64d ", v[i][j]);
			puts("");
		}
		puts("");
	}
} a, b, c;

Matrix operator * (Matrix a, Matrix b) {
	Matrix c(a.n);
	for (int i = 0; i < a.n; i++) {
		for (int j = 0; j < a.n; j++) {
			c.v[i][j] = 0;
			for (int k = 0; k < a.n; k++) {
				c.v[i][j] += (a.v[i][k] * b.v[k][j]) % mod;
				c.v[i][j] %= mod;
			}
		}
	}
	return c;
}

Matrix operator ^ (Matrix a, ll k) {
	Matrix c(a.n);
	c.init(1);
	while (k) {
		if (k & 1)
			c = a * c;
		a = a * a;
		k >>= 1;
	}
	return c;
}

Matrix operator + (Matrix a, Matrix b) {
	Matrix c(a.n);
	for (int i = 0; i < a.n; i++)
		for (int j = 0; j < a.n; j++)
		c.v[i][j] = (b.v[i][j] + a.v[i][j]) % mod;
	return c;
}

Matrix operator + (Matrix a, ll b) {
	Matrix c = a;
	for (int i = 0; i < a.n; i++)
		c.v[i][i] = (a.v[i][i] + b) % mod;
	return c;
}


ll n, m, con[maxn];
ll f[maxn], f2[maxn];

int main() {
	while (scanf("%I64d%I64d", &n, &m) != EOF) {
		for (int i = 1; i <= n; i++)
			scanf("%I64d", &con[i]);

		memset(f, 0, sizeof(f));
		f[0] = 233;
		f[1] = con[1];
		for (int i = 2; i <= n; i++) {
			memcpy(f2, f, sizeof(f));
			for (int j = 1; j < 10; j++)
				f[j] = f2[j] + f2[j-1];
			f[i] = con[i];
			f[0] = f[0] * 10 + 3;
		}

		a.init();
		a.n = n + 2;
		a.v[0][0] = 10;
		a.v[0][n+1] = 1;
		for (int i = 1; i <= n; i++)
			a.v[i][i-1] = a.v[i][i] = 1;
		a.v[n+1][n+1] = 1;

		b.init();
		b.n = n+2;
		for (int i = 0; i <= n; i++)
			b.v[i][0] = f[i];
		b.v[n+1][0] = 3;

		c = a ^ m;
		c = c * b;
		printf("%I64d\n", c.v[n][0]);
	}
	return 0;
}