Blue Bridge: improve the algorithm to take candy

 To improve the algorithm to take candy   

Problem Description

　　Mom to buy a small B N pieces of candy! But she is not allowed to eat small B directly.
　　Suppose there are M pieces of candy current, each B can take a small candy P, wherein P M is not larger than a root of M prime factors. In this case, the mother will be a small piece of candy B later took P P away from sugar candy pile. Then B can then take a small sugar.
　　Up to now small B would like to know how much sugar can get.

Input Format

　　An integer N

Output Format

　　Up to how much sugar can take

Sample input

15

Sample Output

6

Scale data and conventions

　　N <= 100000

Analysis: Dynamic programming problem ~ ~

Therefore especially before it - does not create ⼀ a full EMPTY zoomed at root, the maximum values ​​MAXN prime table, and then the screen for a prime table ⾥ number one by traversing ~

Construction ⼀ a DP [i] array representing the maximum number when the number of confectionery candies could get by when i ~

For values ​​DP [i]: Since ⼩ B can not get below approximately every prime factor in the root number i,

Primes full table traversing prime condition EMPTY (prime [j] <= sqrt (i) && i% prime [j] == 0),

Update dp [i] values ​​(dp [i-2 * prime [j]] + prime [j]), the maximum values ​​of ~

即：dp[i] =max(dp[i], dp[i-2*prime[j]] + prime[j]);

#include <iostream>
#include <cmath>
using namespace std;
int prime[50000];
int dp[100005];
int book[100005];
int cnt = 0;
void create() {
 int len = sqrt(100005);
 for(int i = 2; i <= len; i++) {
     if(book[i] == 0) {
        prime[cnt++] = i;
        for(int j = i * i; j <= len; j = j + i)
            book[j] = 1;
     }
    }
}
int main() {
 create();
 int n;
 cin >> n;
 for(int i = 1; i <= n; i++) {
     for(int j = 0; j < cnt; j++) {
         if( prime[j] > sqrt(i))
             break;
         if( i % prime[j] == 0)
             dp[i] = max(dp[i], dp[i-2*prime[j]] + prime[j]);
     }
 }
 cout << dp[n];
 return 0;
}