Codeforces Round #511 (Div. 1), problem: (A) Enlarge GCD
题意:删除最小的数字使得所有数的gcd比原来的gcd大,无方案输出-1
想法:只要找到大于原先gcd的,并且n个数包含它的倍数最多的就可以了。
做法:原先写了一个分解定理,每个数都分解一遍最后TLE了,只能被迫看题解,不然不能睡觉了
题解这么分解的好处就是不会重复计算,是4的倍数的数肯定是2的倍数,就直接vis判断一下就可以了,真的神奇
是真的太菜了啊,连A都做不出,我真的是0.0
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<queue>
#include<map>
#define ll long long
#define pb push_back
#define rep(x,a,b) for (int x=a;x<=b;x++)
#define repp(x,a,b) for (int x=a;x<b;x++)
#define W(x) printf("%d\n",x)
#define WW(x) printf("%lld\n",x)
#define pi 3.14159265358979323846
#define mem(a,x) memset(a,x,sizeof a)
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
using namespace std;
const int maxn=15*1e6+7;
const int maxn2=3e5+7;
const int INF=1e9;
const ll INFF=1e18;
int a[maxn2];
int vis[maxn];
int M[maxn];
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int main()
{
int n,g,maxx=0;
scanf("%d",&n);
rep(i,1,n)
{
scanf("%d",&a[i]);
if (i==1) g=a[i];
else g=gcd(g,a[i]);
}
rep(i,1,n){a[i]/=g;M[a[i]]++;maxx=max(maxx,a[i]);}
int ans=INF;
for (int i=2;i<=maxx;i++)
{
int res=0;
if (!vis[i])
{
for (int j=i;j<=maxx;j+=i)
{
vis[j]=1;
res+=M[j];
}
}
ans=min(ans,n-res);
}
if (ans==INF)W(-1);
else W(ans);
return 0;
}