POJ 3421分解质因数

X-factor Chains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7375		Accepted: 2340

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 = 2000000;
int prime[MAXN+1];
void getPrime()
{
    memset(prime,0,sizeof(prime));
    for(int i = 2;i <= MAXN;i++)
    {
        if(!prime[i])prime[++prime[0]] = i;
        for(int j = 1;j <= prime[0] && prime[j] <= MAXN/i;j++)
        {
            prime[prime[j]*i] = 1;
            if(i % prime[j] == 0)break;
        }
    }
}
int factor[100][2];//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 = 0;
    long long tmp = x;
    for(int i = 1; prime[i] <= tmp/prime[i];i++)
    {
        factor[fatCnt][1] = 0;
        if(tmp % prime[i] == 0 )
        {
            factor[fatCnt][0] = prime[i];
            while(tmp % prime[i] == 0)
            {
                factor[fatCnt][1] ++;
                tmp /= prime[i];
            }
            fatCnt++;
        }
    }
    if(tmp != 1)
    {
        factor[fatCnt][0] = tmp;
        factor[fatCnt++][1] = 1;
    }
    return fatCnt;
}
ll jc(int x){
    ll s=1;
    for(int i=2;i<=x;i++)
        s*=i;
    return s;
}
int main()
{ 
    getPrime();
    int x;
    while(scanf("%d",&x)==1){
        getFactors(x);
        int ans1=0;
        ll tmp=1;
        for(int i=0;i<fatCnt;i++){
            //cout<<factor[i][0]<<" "<<factor[i][1]<<endl;
            ans1+=factor[i][1];
            tmp*=jc(factor[i][1]);
        }
        printf("%d %lld\n",ans1,jc(ans1)/tmp);
    }
    return 0;
}