abc233_c Product 题解

合集 - 题解(48)

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6.abc233_c Product 题解2023-03-01

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收起

Product

题意

有 $n$ 个集合，第 $i$ 个集合中有 $l_i$ 个数，分别为 $a_{i,1},a_{i,2}\dots a_{i,l_i}$

从每个集合中选取一个数，问：有多少种取法使得取出的数的乘积为 $x$。

数据范围

• $n,l_i⩾2$
• $\prod _{i=1}^n{l}_i⩽{10}^5$
• $1⩽a_{i,j}⩽{10}^9,1⩽x⩽{10}^{18}$

思路

这题其实很easy，只需要爆搜就可以AC了。

可以分析出 $n$ 是不超过 $20$ 的，因为 $l_i$ 至少为 $2$ 且乘积不超过 ${10}^5$，所以 $n={\log }_2\prod _{i=1}^n{l}_i$。

由于题目给的数据范围限制了 $l_i$ 的乘积是不超过 ${10}^5$ 的，所以可以A掉。

复杂度

时间:$O(\prod _{i=1}^n{l}_i)$
空间:$O(n\times \prod _{i=1}^n{l}_i)$

Code

点击查看代码

#include <iostream>
#include <vector>
#include <queue>
 
using namespace std;
using ll = long long;
 
const int L = 1e5 + 10;
 
int n, ans, l[20], a[20][L];
ll x;
 
void dfs (int t, ll y) {
  if (y > x || y < 1) {
    return ;
  }
  if (t == n + 1) {
    if (y == x) {
      ans++;
    }
    return ;
  }
  for (int i = 1; i <= l[t]; i++) {
    dfs(t + 1, y * a[t][i]);
  }
}
 
int main(){
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> x;
  for (int i = 1; i <= n; i++) {
    cin >> l[i];
    for (int j = 1; j <= l[i]; j++) {
      cin >> a[i][j];
    }
  }
  dfs(1, 1);
  cout << ans;
  return 0;
}