233 Matrix（矩陣快速冪）

233 Matrix

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

思路：
題目告訴 a[i][j]=a[i-1][j]+a[i][j-1] 的關係，然後計算出矩陣，帶入板子。

完整代碼：

#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
const ll MOD=10000007;
#define mod(x) ((x)%MOD)
#define maxn 15
ll n,nn;
struct mat		//矩陣
{
    ll m[maxn][maxn];
    mat()
    {
        memset(m,0,sizeof(m));		//記得清空
    }
};
//矩陣乘法
mat mat_mul(mat b,mat a)            //要是左乘我之前錯了好幾次
{
    mat ret;
    for(int i=1;i<=n+2;i++)
    {
        for(int j=1;j<=n+2;j++)
        {
            for(int k=1;k<=n+2;k++)
            {
                ret.m[i][j]=mod(ret.m[i][j]+b.m[i][k]*a.m[k][j]);
            }
        }
    }
    return ret;
}
//矩陣快速冪
mat pow_mat(mat a,ll x)		
{
    mat ret;
    for(int i=1;i<=n+2;i++)
    {
        ret.m[i][i]=1;
    }
    while(x)
    {
        if(x&1)
        {
            ret=mat_mul(ret,a);
        }
        a=mat_mul(a,a);
        x>>=1;
    }
    return ret;
}
/**
    板子和其他的都相似就是左乘不一樣
 */
int main()
{
    while(cin>>n>>nn)
    {
        mat a,b;
        a.m[1][1]=23;
        for(int i=1;i<=n;i++)
        {
            cin>>a.m[i+1][1];
        }
        a.m[n+2][1]=3;
        for(int i=1;i<=n+1;i++)
        {
            b.m[i][1]=10;
        }
        for(int i=1;i<=n+2;i++)
        {
            b.m[i][n+2]=1;
        }
        for(int i=1;i<n+2;i++)
        {
            for(int j=2;j<=i;j++)
            {
                b.m[i][j]=1;
            }
        }
        b=pow_mat(b,nn);
        a=mat_mul(b,a);
        cout<<a.m[n+1][1]<<endl;
    }
    return 0;
}