Codeforces Round #511 (Div. 2) C. Enlarge GCD
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2022-05-09 16:16:46
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http://codeforces.com/contest/1047/problem/C
复杂度很玄学的一个题,不是吗
998ms应该是最极限的了吧。
题目就是给n个数字问最少删掉几个能使得gcd变大。
被辉神一眼识破的题,
问题可以转化成n个数字先除以全部的gcd,那么现在的gcd肯定是1,对吧,就问最少删除几个可以使得gcd不是1.
筛质数,然后对于每个质因子,判断有多少个数字存在这个质因子,取出现数量最大的那么质因子,然后用n-最大出现次数就可以了
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int MAXN=3e5+5;
const int MAXM=1.5e7+5;
const int mmax=1.5e7;
int C[MAXN],num[MAXM],vis[MAXM];
int main()
{
int n;
scanf("%d",&n);
int now=0;
for(int i=1;i<=n;i++){
scanf("%d",&C[i]);
now=__gcd(now,C[i]);
}
int num1=0;
//cout<<now<<endl;
int Max=-1;
for(int i=1;i<=n;i++){
C[i]/=now;
Max=max(Max,C[i]);
if(C[i]==1)num1++;
num[C[i]]++;
}
if(num1==n){
printf("-1\n");return 0;
}
int ans=0;
for(int i=2;i<=Max;i++){
if(vis[i])continue;
int t=0;
for(int j=1;j*i<=Max;j++){
vis[i*j]=1;t+=num[i*j];
}
ans=max(ans,t);
}
printf("%d\n",n-ans);
}
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