Luogu P3303 【[SDOI2013]淘金】

Description

传送门


Solution

首先每一位上有\(0\)就不合法,然后很多不同的排列乘起来是相同的,所以盲猜最后的\(f\)函数有用的值不多。

经过搜索发现最多只有不到\(10000\)个。

为啥别的题解都是8282我搜出来是9200?

那就把这些合法的\(f\)函数值都放进哈希表存起来,然后设\(dp_{i, x, 0/1}\)表示当前\(dp\)到了第\(i\)位,乘积为哈希表里的第\(x\)个状态,有没有卡到最高位。

这个朴素的数位\(dp\)就枚举一下上一位的乘积,这一位选什么数即可。

注意要考虑前导零的情况。

然后我们就搜出来每种合法状态有多少个数会变成它,那么对于一个合法的位置\((i, j)\),它上面就有\(tot_i \times tot_j\)个金矿。

那么对\(tot\)排序,将一个二元组\((i, j)\)放进堆里表示这是\(tot\)数组的第\(i\)个和第\(j\)个数相乘。

\((i, j)\)可以转移到\((i + 1, j)\)\((i, j + 1)\),但是这样直接转移会有一个小问题,\((i - 1, j)\)\((i, j - 1)\)都能转移到\((i, j)\),所以再记录一下上一次是不是将\(j\)加一,如果上次将\(j\)加一,那么这次不能再将\(i\)加一。

这样堆中的状态变为了一个三元组\((i, j, p)\),代表\(tot\)数组的第\(i\)个和第\(j\)个数相乘,上次是不是将\(j\)加一。


Code

/*
    _______                       ________                        _______
   / _____ \                     / ______ \                      / _____ \
  / /     \_\  _     __     _   / /      \ \   _     __     _   / /     \_\
 | |          | |   |  |   | | | |        | | | |   |  |   | | | |
 | |          | |   |  |   | | | |     __ | | | |   |  |   | | | |
 | |       __ \  \  |  |  /  / | |     \ \| | \  \  |  |  /  / | |       __
  \ \_____/ /  \  \/ /\ \/  /   \ \_____\  /   \  \/ /\ \/  /   \ \_____/ /
   \_______/    \___/  \___/     \______/\__\   \___/  \___/     \_______/
*/
#include <iostream>
#include <queue>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

#define ll long long
#define RE register

const int HashMOD = 100077;
const int MOD = 1e9 + 7;

int k, head[HashMOD + 5], num, fa[20], shu[20], len, tot[10005], ans;

ll dp[14][10005][2];

ll n;

struct Node {
	int next;
	ll to;
} edge[HashMOD + 5];

struct HeapNode {
	int pos1, pos2;
	ll qz;
	bool flag;
	bool operator < (const HeapNode &rhs) const {
		return qz < rhs.qz;
	}
};

inline void Add(ll x) {
	int pos = x % HashMOD;
	for (RE int i = head[pos]; i; i = edge[i].next) {
		ll v = edge[i].to;
		if (v == x) return;
	}
	edge[++num] = (Node){head[pos], x};
	head[pos] = num;
	return;
}

inline int Find(ll x) {
	int pos = x % HashMOD;
	for (RE int i = head[pos]; i; i = edge[i].next) {
		ll v = edge[i].to;
		if (v == x) return i;
	} 
	return 0;
}

inline ll Ksm(ll a, int b) {
	ll tmp = 1;
	while (b) {
		if (b & 1) tmp = tmp * a;
		a = a * a;
		b >>= 1;
	}
	return tmp;
}

void Work() {
	ll tmp = 1;
	for (RE ll i = 1; i <= 9; i++)
		tmp *= Ksm(i, fa[i]);
	Add(tmp);
	return;
}

void Dfs(int x, int sum) {
	if (x == 10 || sum == 12) {
		Work();
		return;
	}
	for (RE int i = 0; i <= 12 - sum; i++)
		fa[x] = i, Dfs(x + 1, sum + i);
	return;
}

void Solve(ll n) {
	ll tmp = n;
	while (n) {
		shu[++len] = n % 10;
		n /= 10;
	}	
	for (RE int i = 1; i <= shu[len]; i++) dp[len][Find(i)][i == shu[len]] = 1;
	for (RE int i = len - 1; i >= 1; i--) {
		for (RE int now = 1; now <= 9; now++)
			dp[i][Find(now)][0]++;
		for (RE int lst = 1; lst <= num; lst++) 
			for (RE int p = 0; p <= 1; p++)
				if (dp[i + 1][lst][p]) {
					for (RE int now = 1; now <= (p ? shu[i] : 9); now++) {
						ll ne = edge[lst].to * now;
						dp[i][Find(ne)][p && (now == shu[i])] += dp[i + 1][lst][p];
					}
				}
	}
	for (RE int i = 1; i <= num; i++)
		if (edge[i].to <= tmp)
			tot[i] += dp[1][i][0] + dp[1][i][1];
	return;
}

bool Cmp(int a, int b) {
	return a > b;
}

void Qm(int &x) {
	x -= MOD;
	x += x >> 31 & MOD;
	return;
}

void Calc(int k) {
	sort(tot + 1, tot + num + 1, Cmp);
	priority_queue<HeapNode> q;
	q.push((HeapNode){1, 1, 1LL * tot[1] * tot[1] % MOD, false});
	while (!q.empty() && k) {
		HeapNode tmp = q.top(); q.pop();
		int pos1 = tmp.pos1, pos2 = tmp.pos2;
		if (!tmp.qz) break;
		k--;
		Qm(ans += tmp.qz % MOD);
		if (pos1 <= num && tmp.flag == false) q.push((HeapNode){pos1 + 1, pos2, 1LL * tot[pos1 + 1] * tot[pos2], false});
		if (pos2 <= num) q.push((HeapNode){pos1, pos2 + 1, 1LL * tot[pos1] * tot[pos2 + 1], true});
	}
	return;
}

int main() {
	Dfs(1, 0);
	scanf("%lld%d", &n, &k);
	Solve(n);
	Calc(k);
	printf("%d", ans);
	return 0;
}
posted @ 2020-11-17 17:47  Tian-Xing  阅读(58)  评论(0编辑  收藏  举报