hdu 4344 Mark the Rope （Miller Rabbin + Pollard rho）

Mark the Rope

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2764    Accepted Submission(s): 916

Problem 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.

 

Input

First line: a positive number T (T<=500) representing the number of test cases
2 to T+1 lines: every line has only a positive number N (N<2 ⁶³) representing the length of rope

 

Output

For every test case, you only need to output K and S separated with a space

 

Sample Input

2 180 198

 

Sample Output

3 18 3 22

 

Author

HIT

 

Source

2012 Multi-University Training Contest 5

 

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题解：Miller Rabbin + Pollard rho

Miller Rabbin 可以在O(s(logn)^3)的时间复杂度内判断一个数是否为素数，有2^(-s)的概率出错

Pollard rho 是基于Miller Rabbin的一种快速分解质因数的做法。

该算法的大致流程是先判断当前数是否为素数(用Miller Rabbin)，如果是直接记录下该质数，直接返回。如果不是就试图找到一个因子(可以不是质因子)，然后对于当前因子p，和n/p分别递归寻找质因子。

对于质因数的寻找，我们采用一种随机化的算法。我们假设要找到的质因子为p，先随机取一个x，然后用x构造y，使p=gcd(x-y,n)如果p不等于1那么就找到了一个质因子，如果找不到我们就不断的调整x,使x=x*x+c(c一般可以随机),直到x==y出现循环，则选取失败。重新选取x，重复上述过程。

#include
#include
#include
#include
#include
#define LL long long 
#define N 10000003
using namespace std;
LL n,mx,cnt,num[N],prime[N],c[N];
LL mul(LL a,LL b,LL p)
{
	LL ans=0; LL base=a%p;
	while (b) {
		if (b&1) ans=(ans+base)%p;
		b>>=1;
		base=(base+base)%p;
	}
	return ans;
}
LL quickpow(LL num,LL x,LL p)
{
	LL ans=1; LL base=num%p;
	while (x) {
		if (x&1) ans=mul(ans,base,p);
		x>>=1;
		base=mul(base,base,p);
	}
	return ans;
}
bool miller_rabbin(LL n)
{
	if (n==2) return true;
	if (!(n&1)) return false;
	LL t=0,a,x,y,u=n-1;
	while (!(u&1)) t++,u>>=1;
	for (int i=0;i<=10;i++) {
		a=rand()*rand()%(n-1)+1;
		x=quickpow(a,u,n);
		for (int j=0;j