Codeforces Round #511 (Div. 2)-C

传送门：http://codeforces.com/contest/1047/problem/C

题意：

　　给定n个数，问最少要去掉几个数，使得剩下的数gcd 大于原来n个数的gcd值。

思路：

　　自己一开始想把每个数的因子都找出来，找到这些因子中出现次数最多且因子大于n个数的最大公约数的，（n - 次数 ）就是答案。但是复杂度是1e9，差那么一点。

自己还是对素数筛理解的不够深。这道题可以枚举素数x，对于每个x，找到所有（a【i】/gcd(all)) 是x倍数的个数，就是一个次数。找这个次数的过程正好与素数筛的过程一致。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>
#include <unordered_map>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)

#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue
#define max3(a,b,c) max(max(a,b),c)



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;       
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 256;
const double esp = 1e-8;
const double PI=acos(-1.0);



template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}


/*-----------------------showtime----------------------*/
            const int maxn = 3e5+9;
            const int maxnum = 1.5e7+9;
            int a[maxn],cnt[maxnum];
            int p[maxn];
            bool prime[maxnum];
            int gcd(int a,int b){
                if(b==0)return a;
                return gcd(b, a%b);
            }

int main(){
            int n,g=0;scanf("%d", &n);
            int ans = 0;
            for(int i=1; i<=n; i++){
                scanf("%d", &a[i]);
                g = gcd(g,a[i]);
            }
            for(int i=1; i<=n; i++)cnt[a[i]/g]++;
            int tot = 0;

            for(int i=2; i<maxnum; i++){
                if(!prime[i]){
                    int x = i;int h = 0;
                    for(int j=i; j<maxnum; j+=i){
                        prime[j] = 1;
                        h += cnt[j]; 
                    }
                    ans = max(ans, h);
                }
            }
            if(!ans)puts("-1");
            else printf("%d
", n-ans);
            return 0;
}