cf div2 236 D

D. Upgrading Array
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

You have an array of positive integers a[1], a[2], ..., a[n] and a set of bad prime numbers b1, b2, ..., bm. The prime numbers that do not occur in the set b are considered good. The beauty of array a is the sum , where function f(s) is determined as follows:

  • f(1) = 0;
  • Let's assume that p is the minimum prime divisor of s. If p is a good prime, then , otherwise .

You are allowed to perform an arbitrary (probably zero) number of operations to improve array a. The operation of improvement is the following sequence of actions:

  • Choose some number r (1 ≤ r ≤ n) and calculate the value g = GCD(a[1], a[2], ..., a[r]).
  • Apply the assignments: , , ..., .

What is the maximum beauty of the array you can get?

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 5000) showing how many numbers are in the array and how many bad prime numbers there are.

The second line contains n space-separated integers a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — array a. The third line contains m space-separated integers b1, b2, ..., bm (2 ≤ b1 < b2 < ... < bm ≤ 109) — the set of bad prime numbers.

Output

Print a single integer — the answer to the problem.

Sample test(s)
input
5 2
4 20 34 10 10
2 5
output
-2
input
4 5
2 4 8 16
3 5 7 11 17
output
10

贪心，从右往左扫gcd，若遇到好的质因子少于坏的质因子则可以更新使答案增加

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <set>
 7 
 8 using namespace std;
 9 
10 #define maxn 40005
11 #define INF (1 << 30)
12 
13 bool prime[maxn];
14 int ele[maxn],a[5005],g[5005],b[5005];
15 int len = 0,n,m,ans = 0;
16 
17 int gcd(int x,int y) {
18         return y ? gcd(y,x % y) : x;
19 }
20 
21 int cal(int x) {
22         int sum = 0;
23         for(int i = 2; i * i <= x; ++i) {
24                 if(x % i == 0) {
25                         int v,pos;
26                         pos = lower_bound(b + 1,b + m + 1,i) - b;
27                         v = b[pos] == i ? -1 : 1;
28                         while(x % i == 0) {
29                                 sum += v;
30                                 x /= i;
31                         }
32                 }
33         }
34 
35         if(x != 1) {
36                 int pos = lower_bound(b + 1,b + m + 1,x) - b;
37                 sum += b[pos] == x ? -1 : 1;
38         }
39         return sum;
40 }
41 
42 void solve() {
43         //printf("cal = %d\n",cal(4));
44         for(int i = 1; i <= n; ++i) {
45                 ans += cal(a[i]);
46         }
47 
48         g[1] = a[1];
49         for(int i = 2; i <= n; ++i) {
50                 g[i] = gcd(g[i - 1],a[i]);
51         }
52 
53         int t = 1;
54         for(int i = n; i >= 1; --i) {
55                 int v;
56                 if((v = cal(g[i] / t)) < 0) {
57                         //printf("g = %d v = %d\n",g[i],v);
58                         ans += (-v) * i;
59                         t *= g[i] / t;
60                 }
61         }
62 
63 
64         printf("%d\n",ans);
65 }
66 int main() {
67 
68         //freopen("sw.in","r",stdin);
69 
70         scanf("%d%d",&n,&m);
71         for(int i = 1; i <= n; ++i) scanf("%d",&a[i]);
72         for(int i = 1; i <= m; ++i) {
73                scanf("%d",&b[i]);
74         }
75         b[m + 1] = INF;
76 
77         solve();
78         return 0;
79 }
View Code

 

posted @ 2014-03-26 00:02  hyx1  阅读(237)  评论(0)  编辑  收藏  举报