HDU 2294 pendant (DP + matrix Rapid power dimensionality reduction)

HDU 2294 pendant (DP + matrix Rapid power dimensionality reduction)

ACM

Address: HDU 2294 pendant

Question:
Tuhao made jewelry for his sister. He had K kinds of pearls, n of each. To show off the wealth, he needed to use each Pearl. Ask him how many pieces of jewelry with a length of [1, N] can be made.

Analysis:
1 ≤ n ≤ 1,000,000,000 is terrible.
First, consider DP. Obviously, you can think of the following:
dp[i][j] = (k-(j-1))*dp[i-1][j-1] + j*dp[i-1][j](DP [I] [J] indicates the number of pendants with I and j types of pearls)
If n is too big, you have to consider optimization. It is too difficult to use a matrix to quickly transfer power.
Reference:

Because N is too large, I is regarded as a "phase" to construct a matrix and quickly transfer it through a matrix.
Set the one-dimensional array (DP [I] [0 ~ J]) if the State is set to F and the transition matrix is INIT (k + 1 matrix), there are: init * f = f'; (f' is a new State)
Set answer = GN = DP [1] [k] + dp [2] [k] +... + dp [N] [k] So there is a matrix:

| 1 0...............0 1 |       |g0|        |g1‘| | 0 1 0...............0 |       |f1|        |f1‘| | 0 k-1 2.............0 |       |f2|        |f2‘| | ..................... |   *   |..|    =   |..‘| | 0...0 k-(j-1) j 0...0 |       |fj|        |fj‘| | ..................... |       |..|        |..‘| | 0...............0 1 k |       |fk|        |fk‘| 

Initialization of this Code:[g0, f1, f2, ..., fk] = {0, k, 0, ..., 0}

Code:

/**  Author:      illuz <iilluzen[at]gmail.com>*  Blog:        http://blog.csdn.net/hcbbt*  File:        2294.cpp*  Create Date: 2014-08-03 16:03:27*  Descripton:   */#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>using namespace std;#define repf(i,a,b) for(int i=(a);i<=(b);i++)typedef long long ll;const int N = 31;const int SIZE = 31;// max size of the matrixconst int MOD = 1234567891;int ans[N];int t, n, k;struct Mat{int n;ll v[SIZE][SIZE];// value of matrixMat(int _n = SIZE) {n = _n;memset(v, 0, sizeof(v));}void init(ll _v) {memset(v, 0, sizeof(v));repf (i, 0, n - 1)v[i][i] = _v;}void output() {repf (i, 0, n - 1) {repf (j, 0, n - 1)printf("%lld ", v[i][j]);puts("");}puts("");}} a, b, c;Mat operator * (Mat a, Mat b) {Mat c(a.n);repf (i, 0, a.n - 1) {repf (j, 0, a.n - 1) {c.v[i][j] = 0;repf (k, 0, a.n - 1) {c.v[i][j] += (a.v[i][k] * b.v[k][j]) % MOD;c.v[i][j] %= MOD;}}}return c;}Mat operator ^ (Mat a, ll k) {Mat c(a.n);c.init(1);while (k) {if (k&1) c = c * a;a = a * a;k >>= 1;}return c;}void init() {scanf("%d%d", &n, &k);a.n = b.n = c.n = k + 1;a.init(0);a.v[0][0] = a.v[0][k] = 1;repf (i, 1, k) {if (i > 1)a.v[i][i - 1] = k - i + 1;a.v[i][i] = i;}b.init(0);b.v[1][0] = k;}void solve() {c = a ^ n;c = c * b;printf("%lld\n", c.v[0][0]);}int main() {scanf("%d", &t);while (t--) {init();solve();}return 0;}