HDU-5015 233 Matrix 矩阵快速幂
HDU-5015 矩阵快速幂模板题
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
**
思路：
**
第一列元素为：
0
a1
a2
a3
a4
转化为：
23
a1
a2
a3
a4
3
则第二列为：
23*10+3
23*10+3+a1
23*10+3+a1+a2
23*10+3+a1+a2+a3
23*10+3+a1+a2+a3+a4
3
根据前后两列的递推关系，有等式可得矩阵A的元素为：
嗯，然后跑矩阵快速幂就好了，这里的递推公式是一列一列的来，这个得理清楚。
#include
#include
#include
#include
#include
#define mt(a) memset(a,0,sizeof(a))
using namespace std;
typedef long long ll;
const ll mod=10000007;
struct node{
ll mp[20][20];
ll r,c;
};
//矩阵乘法
node mul(node a,node b){
ll r=a.r;
ll c=b.c;
node temp;
temp.r=r;
temp.c=c;
for(int i=0;i>temp[i];
if(n==0){
cout<