HDU - 5015 - 233 Matrix【思维+矩阵快速幂】

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

题意： 题目规定一个233矩阵，就是第一行依次是0,233,2333,23333…..
给你矩阵的第一列的n个数，让你求在矩阵中点（n，m）坐标的值 mod 1e7，范围： n≤10,m≤109

分析： 根据这个提示我们可以直观的构造模型，因为我们得构造233这个数列，我们只需要两个元素就可以构造即：23和3 F[n+1] = F[n]*10+3即可，我们还知道矩阵不超过10行，所以我们开一个13*13的矩阵即可
第一列 * [矩阵] = 第二列
初始值依次为 23    a(1,0)   a(2,0)   ....    a(n,0)    3
具体见下图（以n = 4为例）

参考代码

#include
#define ll long long
#define mod(x) ((x)%MOD)
using namespace std;

const int MOD = 1e7 + 7;

struct mat {
    ll m[13][13];
}a,unit,ans;
int mm,n;

void init() {
    memset(unit.m,0,sizeof(unit.m));
    memset(a.m,0,sizeof(a.m));
    for(int i = 1;i <= n;i++) {
        cin>>unit.m[0][i];
        unit.m[0][i] = mod(unit.m[0][i]);
    }
    unit.m[0][0] = 23;
    unit.m[0][n+1] = 3;
    for(int i = 0;i <= n;i++) {
        a.m[0][i] = 10;
    }
    for(int i = 1;i <= n;i++)
        for(int j = i;j <= n;j++)
            a.m[i][j] = 1;
    for(int i = 0;i <= n+1;i++) {
        a.m[n+1][i] = 1;
    }

}

mat operator *(mat m1,mat m2) {
    mat t;
    ll r;
    for(int i = 0;i < n+2;i++) {
        for(int j = 0;j < n+2;j++) {
            r = 0;
            for(int k = 0;k < n+2;k++) {
                r = mod(r*1ll + mod(mod(m1.m[i][k]) * 1ll * mod(m2.m[k][j])));
            }
            t.m[i][j] = r;
        }
    }
    return t;
}

mat quick_pow(ll x) {
    mat t = unit;
    while(x) {
        if(x & 1) t = t*a;
        a = a*a;
        x >>= 1;
    }
    return t;
}

int main(){
    ios_base::sync_with_stdio(0);
    while(cin>>n>>mm) {
        init();
        if(mm == 0 && n == 0) {
            cout<<0<continue;
        }
        ans = quick_pow(mm);
        cout<0][n]<return 0;
}

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