Prime Game (prime factorization of sieve + + linear thinking)

Meaning of the questions:

　　Is expressed with a word, the number of distinct prime factors all subranges

Ideas:

　　Prime factor decomposition Needless to say, write down the location of each contribution prime factors (in which occurrences), plus each time interval containing this point and lose the nearest part of the same prime factors overlap before (that is, the contribution of these) .

Code:

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 vector <long long> ve[1000005];
 6 
 7 void add(long long pos, long long x)
 8 {
 9     long long i;
10     for(i=2;i*i<=x;i++) //t
11     {
12         if(x%i==0)
13         {
14             ve[i].push_back(pos);
15             while(x%i==0)
16             {
17                 x /= i;
18             }
19         }
20     }
21     if(x>1) ve[x].push_back(pos);
22 }
23 
24 int main()
25 {
26     long long n, i, j, x;
27     long long sum;
28 
29     for(i=0;i<1000002;i++)
30     {
31         ve[i].push_back(0);
32     }
33 
34     scanf("%lld", &n);
35     for(i=1;i<=n;i++)
36     {
37         scanf("%lld", &x);
38         add(i, x);
39     }
40     sum = 0;
41     for(i=2;i<1000002;i++)
42     {
43         for(j=1; j<ve[i].size(); j++)
44         {
45             sum += (ve[i][j] - ve[i][j-1]) * (n - ve[i][j] + 1);
46         }
47     }
48     printf("%lld", sum);
49     return 0;
50 }

There is one that one, this title T for a night due to add in cycling conditions beginning with the sqrt () function, convinced!

Decomposition of the quality factor is:

 1 #include <stdio.h>
 2 #include <string.h>
 3 
 4 int main()
 5 {
 6     int n, top, i;
 7     int re[100005];
 8     scanf("%d", &n);
 9     top = 0;
10     for(i=2;i*i<=n;i++)
11     {
12         if(n%i==0)
13         {
14             re[top++] = i;
15             while(n%i==0)
16             {
17                 n /= i;
18             }
19         }
20     }
21     if(n>1) re[top++] = n;
22     printf("top = %d\n", top);
23     for(i=0;i<top;i++)
24     {
25         if(i==top-1) printf("%d\n", re[i]);
26         else printf("%d ", re[i]);
27     }
28     return 0;
29 }

This question screen by linear optimization, only ran primes

Code:

 1 #include <bits/stdc++.h>
 2 #define maxn 1e6+2
 3 using namespace std;
 4 
 5 vector <long long> ve[1000005];
 6 long long top;
 7 long long IsPrime[1000005], prime[100005];
 8 
 9 void init()
10 {
11     long long i;
12     for(i=0;i<maxn;i++)
13     {
14         ve[i].push_back(0);
15     }
16 }
17 
18 void GetPrime()
19 {
20     long long i, j;
21     top = 0;
22     memset(IsPrime, 0, sizeof(IsPrime));
23     IsPrime[0] = 1;
24     IsPrime[1] = 1;
25     for(i=2;i<maxn;i++)
26     {
27         if(IsPrime[i]==0) prime[top++] = i;
28         for(j=0;j<top&&i*prime[j]<maxn;j++)
29         {
30             IsPrime[i*prime[j]] = 1;
31             if(i%prime[j]==0) break;
32         }
33     }
34 }
35 
36 void add(long long pos, long long x)
37 {
38     long long i;
39     for(i=0;i<top&&prime[i]*prime[i]<=x;i++)
40     {
41         if(x%prime[i]==0)
42         {
43             ve[prime[i]].push_back(pos);
44             while(x%prime[i]==0)
45             {
46                 x /= prime[i];
47             }
48         }
49     }
50     if(x>1) ve[x].push_back(pos);
51 }
52 
53 int main()
54 {
55     long long n, i, j, x;
56     long long sum;
57 
58     init();
59     GetPrime();
60 
61     scanf("%lld", &n);
62     for(i=1;i<=n;i++)
63     {
64         scanf("%lld", &x);
65         add(i, x);
66     }
67     sum = 0;
68     for(i=2;i<maxn;i++)
69     {
70         for(j=1; j<ve[i].size(); j++)
71         {
72             sum = sum + (ve[i][j] - ve[i][j-1]) * (n - ve[i][j] + 1);
73         }
74     }
75     printf("%lld", sum);
76     return 0;
77 }