POJ 3421

X-factor Chains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5111		Accepted: 1622

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

 

找出x的所有质因子表达式，最大长度就是各质因子指数的和，然后按照排列组合计算即可。

 

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

using namespace std;

#define maxn (1 << 20) + 5

typedef long long ll;

int x,len = 0;
bool prime[maxn];
int ele[100000];

ll cal(int x) {
        ll ans = 1;
        for(int i = 1; i <= x; i++) ans *= i;

        return ans;
}

void init() {
        for(int i = 2; i <= maxn - 5; i++) {
                prime[i] = i % 2 || i == 2;
        }

        for(int i = 2; i * i <= maxn - 5; i++) {
                if(!prime[i]) continue;
                for(int j = i; j * i <= maxn - 5; j++) {
                        prime[j * i] = 0;
                }
        }

        len = 0;
        for(int i = 2; i <= maxn - 5; i++) {
                if(prime[i])
                ele[len++] = i;
        }


}
int main() {
        //freopen("sw.in","r",stdin);

        init();

        while(~scanf("%d",&x)) {
                if(prime[x]) {
                        printf("1 1
");
                } else {
                        int ans1 = 0,len2 = 0;
                        ll ans2 = 0;
                        int num[30];
                        for(int i = 0; i < len && ele[i] <= x && x != 1; i++) {
                                int sum = 0;
                                if(x % ele[i] == 0) {
                                        while(x % ele[i] == 0) {
                                                ans1++;
                                                sum++;
                                                x /= ele[i];
                                        }
                                        num[len2++] = sum;
                                }


                        }

                        ans2 = cal(ans1);
                        for(int i = 0; i < len2; i++)  {
                                ans2 /= cal(num[i]);
                        }
                        printf("%d %I64d
",ans1,ans2);

                }

        }

        return 0;
}