poj3421 X-factor Chains

X-factor Chains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7733		Accepted: 2447

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

大致题意：构造一个序列，使得ai整除ai+1,a0 = 1,an = x,现在给定x,求这样的序列最长的长度是多少？有多少个？

分析：将x分解质因数.从第i个数变成第i-1个数就是去掉一个质因子，那么把x的质因子的次数加上，这就是第一问的答案.第二问实际上就是要求排列数，相同的数在不同的位置只能算一次.那么利用可重复的排列数公式计算即可.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;
ll n,ans,jie[30],sum,yinzi[30],tot,mul;

int main()
{
    jie[0] = 1;
    for (int i = 1; i <= 20; i++)
        jie[i] = jie[i - 1] * i;
    while (scanf("%lld",&n) != EOF)
    {
        tot = ans = sum = 0;
        mul = 1;
        memset(yinzi,0,sizeof(yinzi));
        for (ll i = 2; i * i <= n; i++)
        {
            if (n % i == 0)
            {
                ++tot;
                while (n % i == 0)
                {
                    n /= i;
                    yinzi[tot]++;
                    sum++;
                }
            }
        }
        if (n > 1)
        {
            ++tot;
            ++sum;
            yinzi[tot]++;
        }
        printf("%lld ",sum);
        ans = jie[sum];
        for (int i = 1; i <= tot; i++)
            mul *= jie[yinzi[i]];
        ans /= mul;
        printf("%lld\n",ans);
    }

    return 0;
}