POJ 3421分解质因数

X-factor Chains

Description

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

1 = X₀, X₁, X₂, …, X_m = X

satisfying

X_i < X_i+1 and X_i | X_i+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题意：1 = X₀, X₁, X₂, …, X_m = X，X0~Xm都是X的因子并且递增，给出X求出最长的链，有几条最长的链。代码：

//最长链就是X的素因子的个数，数量就是这些素因子的排列组合(重复的只算一个)
//(全部质因子个数的阶乘)/(每个质因子个数的阶乘)
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int MAXN = ;
int prime[MAXN+];
void getPrime()
{
    memset(prime,,sizeof(prime));
    for(int i = ;i <= MAXN;i++)
    {
        if(!prime[i])prime[++prime[]] = i;
        for(int j = ;j <= prime[] && prime[j] <= MAXN/i;j++)
        {
            prime[prime[j]*i] = ;
            if(i % prime[j] == )break;
        }
    }
}
int factor[][];//factor[i][0]存素因子，factor[i][1]存素因子的个数
int fatCnt;//不重复的素因子个数
int getFactors(long long x)
{
    fatCnt = ;
    long long tmp = x;
    for(int i = ; prime[i] <= tmp/prime[i];i++)
    {
        factor[fatCnt][] = ;
        if(tmp % prime[i] ==  )
        {
            factor[fatCnt][] = prime[i];
            while(tmp % prime[i] == )
            {
                factor[fatCnt][] ++;
                tmp /= prime[i];
            }
            fatCnt++;
        }
    }
    if(tmp != )
    {
        factor[fatCnt][] = tmp;
        factor[fatCnt++][] = ;
    }
    return fatCnt;
}
ll jc(int x){
    ll s=;
    for(int i=;i<=x;i++)
        s*=i;
    return s;
}
int main()
{
    getPrime();
    int x;
    while(scanf("%d",&x)==){
        getFactors(x);
        int ans1=;
        ll tmp=;
        for(int i=;i<fatCnt;i++){
            //cout<<factor[i][0]<<" "<<factor[i][1]<<endl;
            ans1+=factor[i][];
            tmp*=jc(factor[i][]);
        }
        printf("%d %lld\n",ans1,jc(ans1)/tmp);
    }
    return ;
}