HDOJ 5015 233 Matrix
构造矩阵+快速幂
233 Matrix
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 344 Accepted Submission(s): 231
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 ≤ 10 9). The second line contains n integers, a 1,0,a 2,0,...,a n,0(0 ≤ a i,0 < 2 31).
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
Source
2014 ACM/ICPC Asia Regional Xi'an Online
#include
#include
#include
#include
using namespace std;
typedef long long int LL;
const LL MOD=10000007LL;
int n,m;
LL a[20];
struct Matrix
{
int x,y;
LL matrix[12][12];
Matrix()
{
x=y=0;
memset(matrix,0,sizeof(matrix));
}
};
void init(Matrix& m)
{
m.x=m.y=n+2;
for(int i=0;i=0;i--)
scanf("%lld",a+i);
a[n]=233; a[n+1]=3;
Matrix M,ED;
init(M);
ED=MatrixQuickPow(M,m-1);
LL ans=0;
for(int i=0;i