ACM学习历程——HDU5015 233 Matrix（矩阵快速幂）（2014陕西网赛）

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 ≤ 10 ⁹). The second line contains n integers, a _1,0,a _2,0,...,a _n,0(0 ≤ a _i,0 < 2 ³¹).

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

这个题目由于m数据范围很大，故不能直接暴力计算。此处采用矩阵乘法，由矩阵乘法可以由每一列得到下一列。然后矩阵的乘法使用快速幂加快计算。

由2333可以由233乘10加3，于是打算构造n+2行的方阵。

大致如下：

10 0 0 0 ……0 1

10 1 0 0 ……0 1

10 1 1 0 ……0 1

……

10 1 1 1 ……1 1

0   0 0 0 ……0 1

而所要求的列矩阵大致如下：

23……3

a 1,0

a 2,0

……

a n,0

3

递推的正确性可以通过计算验证

此处矩阵通过结构体，运算符重载完成。

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <string>
#define inf 0x3fffffff
#define esp 1e-10
#define N 10000007
#define LL long long

using namespace std;

struct Mat
{
    LL val[15][15];
    int len;

    Mat operator = (const Mat& a)
    {
        for (int i = 0; i < len; ++i)
            for (int j = 0; j < len; ++j)
                val[i][j] = a.val[i][j];
        len = a.len;
        return *this;
    }

    Mat operator * (const Mat& a)
    {
        Mat x;
        memset(x.val, 0, sizeof(x.val));
        x.len = len;
        for (int i = 0; i < len; ++i)
            for (int j = 0; j < len; ++j)
                for (int k = 0; k < len; ++k)
                    if (val[i][k] && a.val[k][j])
                        x.val[i][j] = (x.val[i][j] + (val[i][k]*a.val[k][j])%N)%N;
        return x;
    }

    Mat operator ^ (const int& a)
    {
        int n = a;
        Mat x, p = *this;
        memset(x.val, 0, sizeof(x.val));
        x.len = len;
        for (int i = 0; i < len; ++i)
            x.val[i][i] = 1;
        while (n)
        {
            if (n & 1)
                x = x * p;
            p = p * p;
            n >>= 1;
        }
        return x;
    }
};

int n, m;
LL a[15], ans;

void Make(Mat &p)
{
    p.len = n + 2;
    memset(p.val, 0, sizeof(p.val));
    for (int i = 0; i <= n; ++i)
        p.val[i][0] = 10;
    for (int i = 0; i <= n+1; ++i)
        p.val[i][n+1] = 1;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= i; ++j)
            p.val[i][j] = 1;
}

int main()
{
    //freopen("test.txt", "r", stdin);
    while (scanf("%d%d", &n, &m) != EOF)
    {
        Mat p;
        Make(p);
        p = p ^ m;
        a[0] = 23;
        a[n+1] = 3;
        for (int i = 1; i <= n; ++i)
            scanf("%I64d", &a[i]);
        ans = 0;
        for (int i = 0; i <= n+1; ++i)
            ans = (ans + (p.val[n][i]*a[i])%N)%N;
        printf("%I64d\n", ans);
    }
    return 0;
}