【蓝桥杯冲冲冲】进阶搜索 Anya and Cubes

蓝桥杯备赛 | 洛谷做题打卡day22

• Anya and Cubes

  题面翻译

  给你 $n$ 个数，$n\le 25$。初始序列为 $a_i,0\le a_i\le 10^9$。

  你有 $k$ 个 $!$，每个 $!$ 可以使序列中的一个数变成 $a_i!$。（同一个数至多变一次）

  例如 $5!=120$。

  求：选出任意个数使他们和的等于 $S$ 的方案数

  输入格式

  第一行 $n,k,S$。

  第二行 $n$ 个数，分别代表 $a_i$。

  输出

  一行一个整数，选出数的合法方案数。

  感谢 @s_a_b_e_r 提供的翻译。

  题目描述

  Anya loves to fold and stick. Today she decided to do just that.

  Anya has $ n $ cubes lying in a line and numbered from $ 1 $ to $ n $ from left to right, with natural numbers written on them. She also has $ k $ stickers with exclamation marks. We know that the number of stickers does not exceed the number of cubes.

  Anya can stick an exclamation mark on the cube and get the factorial of the number written on the cube. For example, if a cube reads $ 5 $, then after the sticking it reads $ 5! $, which equals $ 120 $.

  You need to help Anya count how many ways there are to choose some of the cubes and stick on some of the chosen cubes at most $ k $ exclamation marks so that the sum of the numbers written on the chosen cubes after the sticking becomes equal to $ S $. Anya can stick at most one exclamation mark on each cube. Can you do it?

  Two ways are considered the same if they have the same set of chosen cubes and the same set of cubes with exclamation marks.

输入格式

The first line of the input contains three space-separated integers $ n $, $ k $ and $ S\ (1\le n\le25,\ 0\le k\le n,\ 1\le S\le10^{16})$ — the number of cubes and the number of stickers that Anya has, and the sum that she needs to get.

The second line contains $ n $ positive integers $ a_{i}\ ( 1\le a_{i}\le10^{9}) $ — the numbers, written on the cubes. The cubes in the input are described in the order from left to right, starting from the first one.

Multiple cubes can contain the same numbers.

输出格式

Output the number of ways to choose some number of cubes and stick exclamation marks on some of them so that the sum of the numbers became equal to the given number $ S $ .

样例 #1

样例输入 #1

2 2 30
4 3

样例输出 #1

1

样例 #2

样例输入 #2

2 2 7
4 3

样例输出 #2

1

样例 #3

样例输入 #3

3 1 1
1 1 1

样例输出 #3

6

提示

In the first sample the only way is to choose both cubes and stick an exclamation mark on each of them.

In the second sample the only way is to choose both cubes but don’t stick an exclamation mark on any of them.

In the third sample it is possible to choose any of the cubes in three ways, and also we may choose to stick or not to stick the exclamation mark on it. So, the total number of ways is six.

题解代码

学会利用新知，自己多试试并尝试积攒一些固定解答方案，debug，以下是题解代码 ~

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll ans[25], res[25], S;
int n, m, maxDep;
bool f;
bool IDDFS(ll p, ll q, int dep)
{if (dep == maxDep){if (p == 1 && q > res[dep - 1] && (q < ans[dep] || !ans[dep])){res[dep] = q;memcpy(ans, res, sizeof(res));return true;}return false;}if (dep == maxDep - 1){for (ll z = (q << 2) / p / p + 1; z < min((S << 1) / p, S * S / q); z++){ll delta = p * p * z * z - (q << 2) * z, s = sqrt(delta);if (s * s != delta || p * z + s & 1)continue;res[dep] = p * z - s >> 1, res[dep + 1] = p * z + s >> 1;if (res[dep + 1] < ans[dep + 1] || !ans[dep + 1]){memcpy(ans, res, sizeof(res));return true;}}return false;}ll l = max(res[dep - 1], (q - 1) / p) + 1LL, r = (maxDep - dep + 1) * q / p;bool flag = false;for (int i = l; i < r; i++){ll tx = p * i - q, ty = q * i, g = __gcd(tx, ty);res[dep] = i;if (IDDFS(tx / g, ty / g, dep + 1))flag = true;}return flag;
}
int main()
{
#ifndef ONLINE_JUDGEfreopen("data.in", "r", stdin);// freopen("data.out", "w", stdout);
#endifscanf("%d%d", &n, &m);while (!f){S = 100, maxDep++;while (S < 1e7 && !f)S = (S << 3) + (S << 1), f = IDDFS(n, m, 1);}for (int i = 1; i <= maxDep; i++)printf("%lld ", ans[i]);return 0;
}

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