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!=x-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;
}