HDU-5015 233 matrix (matrix Construction)

Problem descriptionin our daily life we often use 233 to express our feelings. actually, we may say 2333,233 33, or 233333... in the same meaning. and here is the question: Suppose we have a matrix called 233 matrix. in the first line, it wocould be 233,233 3, 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, cocould you tell me an, m in the 233 matrix?
Inputthere 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 ).
Outputfor each case, output an, M mod 10000007.
Sample Input

1 112 20 03 723 47 16

 
Sample output

234279972937Hint
  

 
Source2014 ACM/ICPC Asia Regional Xi 'an online
Question: calculate the value of the last digit.
Train of Thought: Move the matrices after line I back to the I bit to construct a matrix:
10 0 0 0 0... 1
1 1 0 0 0... 0
0 1 1 0... 0
................
0 0 0 0 0... 1,
However, an initial matrix with the B dimension n + 2*1 needs to be released, which is an initial Column Based on the movement.

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>typedef __int64 ll;using namespace std;const int maxn = 20;const int mod = 10000007;struct Matrix{int n;ll v[maxn][maxn];Matrix(int _n = maxn) {n = _n;}void init(ll _v = 0) {memset(v, 0, sizeof(v));if (_v)for (int i = 0; i < n; i++)v[i][i] = _v;}void output() {for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++)printf("%I64d ", v[i][j]);puts("");}puts("");}} a, b, c;Matrix operator * (Matrix a, Matrix b) {Matrix c(a.n);for (int i = 0; i < a.n; i++) {for (int j = 0; j < a.n; j++) {c.v[i][j] = 0;for (int k = 0; k < a.n; k++) {c.v[i][j] += (a.v[i][k] * b.v[k][j]) % mod;c.v[i][j] %= mod;}}}return c;}Matrix operator ^ (Matrix a, ll k) {Matrix c(a.n);c.init(1);while (k) {if (k & 1)c = a * c;a = a * a;k >>= 1;}return c;}Matrix operator + (Matrix a, Matrix b) {Matrix c(a.n);for (int i = 0; i < a.n; i++)for (int j = 0; j < a.n; j++)c.v[i][j] = (b.v[i][j] + a.v[i][j]) % mod;return c;}Matrix operator + (Matrix a, ll b) {Matrix c = a;for (int i = 0; i < a.n; i++)c.v[i][i] = (a.v[i][i] + b) % mod;return c;}ll n, m, con[maxn];ll f[maxn], f2[maxn];int main() {while (scanf("%I64d%I64d", &n, &m) != EOF) {for (int i = 1; i <= n; i++)scanf("%I64d", &con[i]);memset(f, 0, sizeof(f));f[0] = 233;f[1] = con[1];for (int i = 2; i <= n; i++) {memcpy(f2, f, sizeof(f));for (int j = 1; j < 10; j++)f[j] = f2[j] + f2[j-1];f[i] = con[i];f[0] = f[0] * 10 + 3;}a.init();a.n = n + 2;a.v[0][0] = 10;a.v[0][n+1] = 1;for (int i = 1; i <= n; i++)a.v[i][i-1] = a.v[i][i] = 1;a.v[n+1][n+1] = 1;b.init();b.n = n+2;for (int i = 0; i <= n; i++)b.v[i][0] = f[i];b.v[n+1][0] = 3;c = a ^ m;c = c * b;printf("%I64d\n", c.v[n][0]);}return 0;}

HDU-5015 233 matrix (matrix Construction)