Codeforces Round #697 (Div. 3)
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比赛链接
Codeforces Round #697 (Div. 3)
G. Strange Beauty
题目大意: 有 \(n\) 个数,从中挑选一个最大的子集,使得集合中任意两个不同的数 \(x, y\) ,有 \(x \mid y\) 或 \(y \mid x\)
输入格式
The first line contains one integer \(t(1 \leq t \leq 10)\) - the number of test cases. Then \(t\) test cases follow.
The first line of each test case contains one integer \(n\left(1 \leq n \leq 2 \cdot 10^{5}\right)-\) the length of the array \(a\)
The second line of each test case contains \(n\) numbers \(a_{1}, a_{2}, \ldots, a_{n}\left(1 \leq a_{i} \leq 2 \cdot 10^{5}\right)-\) elements of the array \(a\).
输出格式
For each test case output one integer - the minimum number of elements that must be removed to make the array \(a\) beautiful.
解题思路
dp
不难发现,值域较小,可将所有出现的值作为桶,统计每个数出现的次数
-
状态表示:\(f[i]\) 表示最大值为 \(i\) 的满足条件的最长上升子序列的长度
-
状态计算:\(f[i]+=a[i]\)
分析:值 \(i\) 作为最大值时,应加上值为 \(i\) 出现的次数,注意这里不是 \(f[i]=a[i]\),因为前面可能会有 \(i\) 的约数也作为最长长度的贡献 -
时间复杂度:\(O(nlogn)\)
代码
// Problem: G. Strange Beauty
// Contest: Codeforces - Codeforces Round #697 (Div. 3)
// URL: https://codeforces.com/contest/1475/problem/G
// Memory Limit: 256 MB
// Time Limit: 5000 ms
//
// Powered by CP Editor (https://cpeditor.org)
// %%%Skyqwq
#include <bits/stdc++.h>
//#define int long long
#define help {cin.tie(NULL); cout.tie(NULL);}
#define pb push_back
#define fi first
#define se second
#define mkp make_pair
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }
template <typename T> void inline read(T &x) {
int f = 1; x = 0; char s = getchar();
while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
x *= f;
}
const int N=2e5+5;
int t,n,f[N],a[N];
int main()
{
for(cin>>t;t;t--)
{
cin>>n;
memset(f,0,sizeof f);
memset(a,0,sizeof a);
for(int i=1;i<=n;i++)
{
int x;
cin>>x;
a[x]++;
}
int res=0;
for(int i=1;i<N;i++)
{
f[i]+=a[i];
res=max(res,f[i]);
for(int j=i;j<N;j+=i)f[j]=max(f[j],f[i]);
}
cout<<n-res<<'\n';
}
return 0;
}
标签:697,int,Codeforces,long,leq,test,Div,define 来源: https://www.cnblogs.com/zyyun/p/16410217.html