【Codeforces403B】【贪心】Upgrading Array

Upgrading Array

Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u

Description

You have an array of positive integers a[1], a[2], …, a[n] and a set of bad prime numbers b1, b2, …, bm. The prime numbers that do not occur in the set b are considered good. The beauty of array a is the sum , where function f(s) is determined as follows:
f(1) = 0;
Let’s assume that p is the minimum prime divisor of s. If p is a good prime, then , otherwise .
You are allowed to perform an arbitrary (probably zero) number of operations to improve array a. The operation of improvement is the following sequence of actions:
Choose some number r (1 ≤ r ≤ n) and calculate the value g = GCD(a[1], a[2], …, a[r]).
Apply the assignments: , , …, .
What is the maximum beauty of the array you can get?

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 5000) showing how many numbers are in the array and how many bad prime numbers there are.
The second line contains n space-separated integers a[1], a[2], …, a[n] (1 ≤ a[i] ≤ 109) — array a. The third line contains m space-separated integers b1, b2, …, bm (2 ≤ b1 < b2 < … < bm ≤ 109) — the set of bad prime numbers.

Output

Print a single integer — the answer to the problem.

Samples

Input1

5 2
4 20 34 10 10
2 5

Output1

-2

Input2

4 5
2 4 8 16
3 5 7 11 17

Output2

10

Hint

Note that the answer to the problem can be negative.
The GCD(x1, x2, …, xk) is the maximum positive integer that divides each xi.

Source

Codeforces Round #236 (Div. 1)

贪心的题，先算出所有的gcd，然后从后向前贪心，如果当前除gcd有贡献就除，不然向前
然后不知道那个题解说的可以暴力分解，我就T了，最后还是加了个筛子

下面直接放代码：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<set>#include<map>#include<queue>#include<algorithm>#include<vector>#include<cstdlib>#include<cmath>#include<ctime>#include<stack>#define INF 2100000000#define ll long long#define clr(x)  memset(x,0,sizeof(x))#define clrmax(x)  memset(x,127,sizeof(x))using namespace std;inline int read(){    char c;    int ret=0;    while(!(c>='0'&&c<='9'))        c=getchar();    while(c>='0'&&c<='9')    {        ret=(c-'0')+(ret<<1)+(ret<<3);        c=getchar();    }    return ret;}#define M 5005int a[M],b[M],n,m,ans,g[M],di=1;int p[M*2],pn,s[M*10];map<int,int>c;void get_prime(){    for(int i=2;i<=240;i++)    {        if(s[i])continue;        for(int j=i+i;j<=50000;j+=i)            s[j]=1;    }    for(int i=2;i<=50000;i++)        if(!s[i])p[++pn]=i;}int gcd(int m,int n){    return n==0?m:gcd(n,m%n);}bool in(int x){    if(x<50000)        if(s[x]==-1)return 1;        else return 0;    else if(c[x])return 1;         else return 0;}int get_gong(int x){    int ret=0;    for(int i=1;i<=pn&&p[i]<=sqrt(x+0.5);i++)    {        while(x%p[i]==0)        {            if(in(p[i]))ret--;            else ret++;            x/=p[i];        }    }    if(x!=1)    {        if(c[x])ret--;        else ret++;    }    return ret;}int main(){    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);    get_prime();    n=read();m=read();    for(int i=1;i<=n;i++)        a[i]=read();    for(int i=1;i<=m;i++)    {        b[i]=read();        if(b[i]<50000)s[b[i]]=-1;        c[b[i]]=1;    }    g[1]=a[1];ans+=get_gong(a[1]);    for(int i=2;i<=n;i++)    {        g[i]=gcd(g[i-1],a[i]);        ans+=get_gong(a[i]);    }    for(int i=n;i>=1;i--)    {        g[i]/=di;        int t=get_gong(g[i]);        if(t<0)        {            ans-=t*i;            di*=g[i];        }    }    cout<<ans;    return 0;}

大概就是这个样子，如果有什么问题，或错误，请在评论区提出，谢谢。