[HDU 4344]Mark the Rope（Pollard_rho+Miller_Rabin）

Description

Eric has a long rope whose length is N, now he wants to mark on the rope with different colors. The way he marks the rope is:
1. He will choose a color that hasn’t been used
2. He will choose a length L (N>L>1) and he defines the mark’s value equals L
3. From the head of the rope, after every L length, he marks on the rope (you can assume the mark’s length is 0 )
4. When he chooses the length L in step 2, he has made sure that if he marks with this length, the last mark will be at the tail of the rope
Eric is a curious boy, he want to choose K kinds of marks. Every two of the marks’ value are coprime(gcd(l1,l2)=1). Now Eric wants to know the max K. After he chooses the max K kinds of marks, he wants to know the max sum of these K kinds of marks’ values.
You can assume that Eric always can find at least one kind of length to mark on the rope.

Solution

质因数分解一下就好了，应该是道模板题

然而我忘记还有质因数只有一个的情况，L不能等于N啊

QvQ拍了好久都没找到错在哪，后来随手输了个1024发现事情有蹊跷…

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;
LL T,ans=0;
vector<LL>v;
map<LL,LL>num;
LL gcd(LL a,LL b)
{
    return b?gcd(b,a%b):a;
}
LL mul(LL a,LL b,LL p)
{
    LL res=0;
    while(b)
    {
        if(b&1)res=(res+a)%p;
        a=(a+a)%p;b>>=1;
    }
    return res;
}
LL pow(LL a,LL n,LL p)
{
    LL res=1;
    while(n)
    {
        if(n&1)res=mul(res,a,p);
        a=mul(a,a,p);n>>=1;
    }
    return res;
}
bool check(LL a,LL n,LL m,LL cnt)
{
    LL x=pow(a,m,n),y=x;
    for(int i=1;i<=cnt;i++)
    {
        x=mul(x,x,n);
        if(x==1&&y!=1&&y!=n-1)return 1;
        y=x;
    }
    return x!=1;
}
bool Miller_Rabin(LL n)
{
    if(n==2)return 1;
    if(n<=1||n&1==0)return 0;
    LL m=n-1,cnt=0;
    while(m&1==0)cnt++,m>>=1;
    for(int i=1;i<=7;i++)
    {
        if(check(rand()%(n-1)+1,n,m,cnt))
        return 0;
    }
    return 1;
}
LL rho(LL n,LL c)
{
    LL i=1,k=2,x=rand()%(n-1)+1,y=x,d;
    while(1)
    {
        x=(mul(x,x,n)+c)%n;
        d=x>y?gcd(x-y,n):gcd(y-x,n);
        if(d>1)return d;
        if(y==x)return n;
        if(i==k)y=x,k<<=1;
        i++;
    }
}
void solve(LL n)
{
    if(n==1)return;
    if(Miller_Rabin(n))
    {
        if(!num[n])
        {v.push_back(n),ans++,num[n]=1;}
        num[n]*=n;
        return;
    }
    LL p=n;
    while(p==n)p=rho(n,rand()%(n-1)+1);
    solve(p);solve(n/p);
}
int main()
{
    scanf("%lld",&T);
    while(T--)
    {
        LL n;
        scanf("%lld",&n);
        v.clear(),num.clear(),ans=0;
        solve(n);
        if(v[0]==n)ans--;
        printf("%lld ",ans);
        LL sum=0;
        for(int i=0;i<ans;i++)
        sum+=num[v[i]];
        if(ans==1)sum/=v[0];
        printf("%lld\n",sum);
    }
    return 0;
}