poj 3421 X-factor Chains

X-factor Chains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5326		Accepted: 1683

Description

Given a positive integer X, an X-factor chain of length m is a sequence of integers,

1 = X0, X1, X2, …, Xm = X

satisfying

Xi < Xi+1 and Xi | Xi+1 where a | b means a perfectly divides into b.

Now we are interested in the maximum length of X-factor chains and the number of chains of such length.

Input

The input consists of several test cases. Each contains a positive integer X (X ≤ 220).

Output

For each test case, output the maximum length and the number of such X-factors chains.

Sample Input

2
3
4
10
100

Sample Output

1 1
1 1
2 1
2 2
4 6

Source

POJ Monthly--2007.10.06, ailyanlu@zsu

分解质因数，然后求一个组合数就成。

s=p1^a1*p2^a2*p3^a3.....pn^an

组合数为：(a1+a2...an)!/(a1!*a2!...*an!)

/*
Problem: 3421		User: motefly
Memory: 3040K		Time: 344MS
Language: G++		Result: Accepted
*/
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

const int maxn=1048600;

int prime[maxn+5];
bool is_prime[maxn+6];
bool hash[maxn+5];

#define ll long long

int sieve(int n)
{
int p=0;
for(int i=0;i<=n;i++)
is_prime[i]=true;
is_prime[0]=is_prime[1]=0;
for(int i=2;i<=n;i++)
{
if(is_prime[i])
{
prime[p++]=i;
hash[i]=1;
for(int j=2*i;j<=n;j+=i)
is_prime[j]=0;
}
}
hash[1]=1;
return p;
}

ll f[21];
int a[21];
void init()
{
sieve(maxn);
f[1]=1;
for(int i=2;i<=20;i++)
f[i]=f[i-1]*i;
}

int main()
{
init();
ll n;
while(~scanf("%lld",&n)){
int k=0;
if(hash
==1)
{
printf("1 1\n");
continue;
}
for(int i=0;prime[i]<=n;i++)
{
if(n==1)
break;
a[k]=0;
if(n%prime[i]==0)
{
while(n%prime[i]==0)
{
n/=prime[i];
a[k]++;
}
k++;
}
}
int sum1=0;
ll sum2=1;
for(int i=0;i<k;i++)
{
sum1+=a[i];
sum2*=f[a[i]];
}
printf("%d %lld\n",sum1,f[sum1]/sum2);
}
}