233 Matrix（矩阵快速幂）

233 Matrix

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 553    Accepted Submission(s): 345

Problem Description

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?

 

 

Input

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).

 

 

Output

For each case, output a_n,m mod 10000007.

 

 

Sample Input

1 1

1

2 2

0 0

3 7

23 47 16

 

 

Sample Output

 

234

2799

72937

Hint

 

       题意：

       有个 N 行 M 列的矩阵，给出第一行和第一列的元素，根据 a [ i ] [ j ] = a [ i - 1 ] [ j ] + a [ i ] [ j - 1 ] 求出 a [ n ] [ m ] 元素是什么。

 

       思路：

       矩阵快速幂。详细题解明天再补回。记得要用 long long。

 

       AC：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

const ll MOD = 10000007;

ll n, m;

mat mul (mat a, mat b) {
    mat c(n + 2, vec(n + 2));

    for (ll i = 0; i < n + 2; ++i) {
        for (ll j = 0; j < n + 2; ++j) {
            for (ll k = 0; k < n + 2; ++k) {
                c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % MOD;
            }
        }
    }

    return c;
}

mat pow (mat a) {
    mat b(n + 2, vec(n + 2));
    for (ll i = 0; i < n + 2; ++i) {
        b[i][i] = 1;
    }

    while (m > 0) {
        if (m & 1) b = mul(b, a);
        a = mul(a, a);
        m >>= 1;
    }

    return b;
}

int main () {

    while (~scanf("%I64d%I64d", &n, &m)) {
        mat a(n + 2, vec(n + 2));
        mat b(n + 2, vec(1));

        a[0][0] = 10, a[0][1] = a[1][1] = 1;
        for (ll i = 2; i < n + 2; ++i) {
            for (ll j = 0; j < n + 2; ++j) {
                if (j <= i) a[i][j] = 1;
                if (!j) a[i][j] = 10;
            }
        }

        b[0][0] = 23, b[1][0] = 3;
        for (ll i = 0; i < n; ++i) {
            scanf("%I64d", &b[i + 2][0]);
            b[i + 2][0] %= MOD;
        }

        a = pow(a);
        b = mul(a, b);

        printf("%I64d\n", b[n + 1][0]);
    }

    return 0;
}