hdu 3092 简单数论+分组背包dp

Least common multiple

Partychen like to do mathematical problems. 
One day, when he was doing on a least common multiple(LCM) problem,
 he suddenly thought of a very interesting question: if given a number of S, 
and we divided S into some numbers , then what is the largest LCM of these numbers? 
partychen thought this problems for a long time but with no result, so he turned to you for help! 
Since the answer can very big,you should give the answer modulo M. 

【题意】给你一个数N和结果的mod M。求如何划分N，使得子集和为N的集合的LCM最大。

从数学上来看，lcm(a, b) = a/gcd(a, b)*b, 只有在gcd(a, b)为1的时候，lcm最大，也就是分割的集合里的每一个数都是互质的。

每个质数和它的幂次作为分组背包的一组（因为每个数互质的时候，lcm最大）；由于dp的时候取模会导致dp结果不正确，而不取模又会导致dp结果过大。由对数的单调性，和对数加法和乘法的性质，dp[i]为i的对数的分割的最大lcm。同时需要另一个数组保留最后结果。

代码如下：

#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
#define ll long long

int prime[3100], pri_size = 0;
bool isprime[3100];
void sieve(int num)
{
    memset(prime, 0, sizeof(prime));
    memset(isprime, true, sizeof(isprime));
    isprime[0] = isprime[1] = false;
    for(int i = 2; i <= num; i++ ) {
        if(isprime[i]) {
            prime[pri_size++] = i;
           // cout << i << " ";
        }
        for(int j = 0; i*prime[j] <= num; j++){
            isprime[i*prime[j]] = false;
            if(i%prime[j] == 0) break;
        } 
    }
   // cout << endl;
}
int main()
{
    int n, m;
    sieve(3000);
    while(cin >> n >> m) {
        int ans[3100]; 
        double dp[3100];
        for(int i = 0; i <= n; i++ ) {dp[i] = 0; ans[i] = 1;}
        
        for(int i = 0; i < pri_size && prime[i] <= n; i++) {
            double log_p = log(prime[i]);
            for(int j = n; j >= 1; j--) {
                ll pri = prime[i], cnt = 1;
                while(pri <= j) {
                    if(dp[j] < dp[j-pri] + log_p*cnt) {
                        dp[j] = dp[j-pri] + log_p * cnt;
                        ans[j] = ans[j-pri] * pri % m;
                    }
                    cnt++;
                    pri *= prime[i];
                }
            }
        }
        cout << ans[n] << endl;
    }
}