POJ 3421 X-factor Chains （约数枚举）

X-factor Chains

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5605 Accepted: 1770

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

23410100

Sample Output

1 11 12 12 24 6

枚举约数 然后dp即可

AC代码如下：

////POJ 3421 X-factor Chains////  Created by TaoSama on 2015-03-31//  Copyright (c) 2015 TaoSama. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int N = 1e5 + 10;int n, len[1500], dp[1500];int main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);//freopen("out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(cin >> n) {        vector<int> factor;        for(int i = 1; i * i <= n; ++i) {            if(n % i == 0) {                factor.push_back(i);                if(i != n / i)  factor.push_back(n / i);            }        }        sort(factor.begin(), factor.end());        int sz = factor.size();        for(int i = 0; i < sz; ++i) {            len[i] = dp[i] = 1;            for(int j = 0; j < i; ++j) {                if(factor[i] % factor[j] == 0) {                    if(len[i] < len[j] + 1) {                        len[i] = len[j] + 1;                        dp[i] = dp[j];                    } else if(len[i] == len[j] + 1)                        dp[i] += dp[j];                }            }        }        cout << len[sz - 1] - 1 << ' ' << dp[sz - 1] << endl;    }    return 0;}