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HDU - 2294 Pendant （DP滚动数组降维+矩阵快速幂）

2024-07-14 21:20:32   217人阅读

Description

On Saint Valentine‘s Day, Alex imagined to present a special pendant to his girl friend made by K kind of pearls. The pendant is actually a string of pearls, and its length is defined as the number of pearls in it. As is known to all, Alex is very rich, and he has N pearls of each kind. Pendant can be told apart according to permutation of its pearls. Now he wants to know how many kind of pendant can he made, with length between 1 and N. Of course, to show his wealth, every kind of pendant must be made of K pearls.
Output the answer taken modulo 1234567891.
 

Input

The input consists of multiple test cases. The first line contains an integer T indicating the number of test cases. Each case is on one line, consisting of two integers N and K, separated by one space.
Technical Specification

1 ≤ T ≤ 10
1 ≤ N ≤ 1,000,000,000
1 ≤ K ≤ 30
 

Output

Output the answer on one line for each test case.
 

Sample Input

2
2 1
3 2
 

Sample Output

2
8

题意：求用k种珍珠组成长度为n的项链的个数

思路：用dp[i][j]表示长度为i，j种珍珠的个数。很容易推出dp[i][j] = dp[i]-1[j]*j+ dp[i-1][j-1]*(k-j+1),因为数据量很大，所以我们需要用矩阵优化，关键构造出矩阵，本来我们是用k维的矩阵构造关系矩阵，但是现在我们要求的是：

  dp[1][k]+dp[1][k]+....dp[n][k]，所以我们都加一维来记录和。

首先我们利用滚动数组降维的思路构造一个矩阵：f[j] = f[j-1]*j + f[j]*(k-j+1), 因为我们需要的是和以及fk，所以第一维就确定下来了

  | 1 0...............0 1  |          |g|

  | 0 1 0...............0  |          |f1|  
  | 0 k-1 2.............0 |          |f2|
  | .....................      |      *   .

  | 0...0 k-(j-1) j 0...0|          .
  | .....................     |           .
  | 0...............0 1 k |           |fk|

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
typedef unsigned long long ll;
const int maxn = 35;
const int mod = 1234567891;

int cnt;
struct Matrix {
	ll v[maxn][maxn];
	Matrix() {}
	Matrix(int x) {
		init();
		for (int i = 0; i < maxn; i++) 
			v[i][i] = x;
	}
	void init() {
		memset(v, 0, sizeof(v));
	}
	Matrix operator *(Matrix const &b) const {
		Matrix c;
		c.init();
		for (int i = 0; i < cnt; i++)
			for (int j = 0; j < cnt; j++)
				for (int k = 0; k < cnt; k++)
					c.v[i][j] = (c.v[i][j] + (ll)(v[i][k]*b.v[k][j])) % mod;
		return c;
	}
	Matrix operator ^(int b) {
		Matrix a = *this, res(1);
		while (b) {
			if (b & 1)
				res = res * a;
			a = a * a;
			b >>= 1;
		}
		return res;
	}
} a, b, tmp;

int main() {
	int t, n, k;
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n, &k);	
		a.init();
		a.v[0][0] = a.v[0][k] = 1;
		for (int j = 1; j <= k; j++) {
			if (j > 1)
				a.v[j][j-1] = k-(j-1);
			a.v[j][j] = j;
		}
		cnt = k + 1;
		ll num[maxn];
		memset(num, 0, sizeof(num));
		num[1] = k;
		tmp = a^n;
		ll ans[maxn];
		memset(ans, 0, sizeof(ans));
		for (int i = 0; i < cnt; i++)
			if (num[i])
				for (int j = 0; j < cnt; j++)
					if (tmp.v[j][i])
						ans[j] = (ans[j]+ (ll)(tmp.v[j][i]*num[i])) % mod;
		cout << ans[0] << endl;
	}
	return 0;
}


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