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：

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 #include 
 6 using namespace std;
 7 #define ll long long
 8 
 9 int prime[3100], pri_size = 0;
10 bool isprime[3100];
11 void sieve(int num)
12 {
13     memset(prime, 0, sizeof(prime));
14     memset(isprime, true, sizeof(isprime));
15     isprime[0] = isprime[1] = false;
16     for(int i = 2; i <= num; i++ ) {
17         if(isprime[i]) {
18             prime[pri_size++] = i;
19            // cout << i << " ";
20         }
21         for(int j = 0; i*prime[j] <= num; j++){
22             isprime[i*prime[j]] = false;
23             if(i%prime[j] == 0) break;
24         } 
25     }
26    // cout << endl;
27 }
28 int main()
29 {
30     int n, m;
31     sieve(3000);
32     while(cin >> n >> m) {
33         int ans[3100]; 
34         double dp[3100];
35         for(int i = 0; i <= n; i++ ) {dp[i] = 0; ans[i] = 1;}
36         
37         for(int i = 0; i < pri_size && prime[i] <= n; i++) {
38             double log_p = log(prime[i]);
39             for(int j = n; j >= 1; j--) {
40                 ll pri = prime[i], cnt = 1;
41                 while(pri <= j) {
42                     if(dp[j] < dp[j-pri] + log_p*cnt) {
43                         dp[j] = dp[j-pri] + log_p * cnt;
44                         ans[j] = ans[j-pri] * pri % m;
45                     }
46                     cnt++;
47                     pri *= prime[i];
48                 }
49             }
50         }
51         cout << ans[n] << endl;
52     }
53 }

View Code

 

 

 

转载于:https://www.cnblogs.com/ZiningTang/p/10478916.html