HDU 5015 233 Matrix

233 Matrix

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1515    Accepted Submission(s): 890


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
a0,1 = 233,a0,2 = 2333,a0,3 = 23333...) Besides, in 233 matrix, we got ai,j = ai-1,j +ai,j-1( i,j ≠ 0). Now you have known a1,0,a2,0,...,an,0, could you tell
me an,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 ≤ 109). The second line contains n integers, a1,0,a2,0,...,an,0(0 ≤ ai,0 < 231).

 

Output

For each case, output an,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 <cstdio>
#include <cstring>

typedef long long LL;
const LL MOD = 10000007;

const int N = 13;

struct matrix{
LL ary

;

void init() {
memset(ary, 0, sizeof(ary));
}

matrix() {
init();
}
};

const matrix operator*(const matrix & A, const matrix & B) {
matrix t;
for (int i = 0; i < N; ++i)
for (int j = 0; j < N; ++j) {
for (int k = 0; k < N; ++k)
t.ary[i][j] += A.ary[i][k] * B.ary[k][j];
t.ary[i][j] = (t.ary[i][j] + MOD) % MOD;
}
return t;
}

LL quick_pow(LL n, LL k) {
matrix ans, tmp;
tmp.ary[0][0] = 10;
tmp.ary[0][n + 1] = tmp.ary[n + 1][n + 1] = 1;
ans.ary[0][0] = 23;
ans.ary[n + 1][0] = 3;
for (int i = 1; i <= n; ++i) {
scanf("%lld", &ans.ary[i][0]);
tmp.ary[i][0] = 10;
tmp.ary[i][n + 1] = 1;
for (int j = 1; j <= i; ++j)
tmp.ary[i][j] = 1;
}

while (k) {
if (k & 1)
ans = tmp * ans;
k >>= 1;
tmp = tmp * tmp;
}

return ans.ary
[0];
}

int main() {
LL n, m;

while (~scanf("%lld%lld", &n, &m))
printf("%lld\n", quick_pow(n, m));
return 0;
}