 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<cstring>
 4 #include<cstdlib>
 5 using namespace std;
 6 
 7 typedef long long LL;
 8 const int maxn = 1e7+1;
 9 bool s[maxn];
10 int prime[maxn],len = 0;
11 int mu[maxn];
12 int g[maxn];
13 int sum1[maxn];
14 void  init()
15 {
16     memset(s,true,sizeof(s));
17     mu[1] = 1;
18     for(int i=2; i<maxn; i++)
19     {
20         if(s[i] == true)
21         {
22             prime[++len]  = i;
23             mu[i] = -1;
24             g[i] = 1;
25         }
26         for(int j=1; j<=len && (long long)prime[j]*i<maxn; j++)
27         {
28             s[i*prime[j]] = false;
29             if(i%prime[j]!=0)
30             {
31                 mu[i*prime[j]] = -mu[i];
32                 g[i*prime[j]] = mu[i] - g[i];
33             }
34             else
35             {
36                 mu[i*prime[j]] = 0;
37                 g[i*prime[j]] = mu[i];
38                 break;
39             }
40         }
41     }
42     for(int i=1; i<maxn; i++)
43         sum1[i] = sum1[i-1]+g[i];
44 }
45 
46 int main()
47 {
48     int a;
49     init();
50     while(scanf("%d",&a)>0)
51     {
52         LL sum = 0;
53         for(int i=1,la = 0 ; i<=a; i = la+1)
54         {
55             la = a/(a/i);
56             sum = sum + (long long)(sum1[la] - sum1[i-1])*(a/i)*(a/i);
57         }
58         printf("%lld\n",sum);
59     }
60     return 0;
61 }

spoj

4491. Primes in GCD Table

Problem code: PGCD

Johnny has created a table which encodes the results of some operation -- a function of two arguments. But instead of a boring multiplication table of the sort you learn by heart at prep-school, he has created a GCD (greatest common divisor) table! So he now has a table (of height a and width b), indexed from (1,1) to (a,b), and with the value of field (i,j) equal to gcd(i,j). He wants to know how many times he has used prime numbers when writing the table.

Input

First, t ≤ 10, the number of test cases. Each test case consists of two integers, 1 ≤ a,b < 10⁷.

Output

For each test case write one number - the number of prime numbers Johnny wrote in that test case.

Example

Input:

2
10 10
100 100

Output:

30
2791

一样的题，只不过 GCD(x,y) = 素数 .  1<=x<=a ; 1<=y<=b;

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<cstring>
 4 #include<cstdlib>
 5 using namespace std;
 6  
 7 typedef long long LL;
 8 const int maxn = 1e7+1;
 9 bool s[maxn];
10 int prime[maxn],len = 0;
11 int mu[maxn];
12 int g[maxn];
13 int sum1[maxn];
14 void  init()
15 {
16     memset(s,true,sizeof(s));
17     mu[1] = 1;
18     for(int i=2;i<maxn;i++)
19     {
20         if(s[i] == true)
21         {
22             prime[++len]  = i;
23             mu[i] = -1;
24             g[i] = 1;
25         }
26         for(int j=1;j<=len && (long long)prime[j]*i<maxn;j++)
27         {
28             s[i*prime[j]] = false;
29             if(i%prime[j]!=0)
30             {
31                 mu[i*prime[j]] = -mu[i];
32                 g[i*prime[j]] = mu[i] - g[i];
33             }
34             else
35             {
36                 mu[i*prime[j]] = 0;
37                 g[i*prime[j]] = mu[i];
38                 break;
39             }
40         }
41     }
42     for(int i=1;i<maxn;i++)
43         sum1[i] = sum1[i-1]+g[i];
44 }
45  
46 int main()
47 {
48     int T,a,b;
49     init();
50     scanf("%d",&T);
51     while(T--)
52     {
53         scanf("%d%d",&a,&b);
54         if(a>b) swap(a,b);
55         LL sum = 0;
56         for(int i=1,la = 0 ;i<=a;i = la+1)
57         {
58             la = min(a/(a/i),b/(b/i));
59             sum = sum + (long long)(sum1[la] - sum1[i-1])*(a/i)*(b/i);
60         }
61         printf("%lld\n",sum);
62     }
63     return 0;
64 }

bzoj 2818: Gcd GCD(a,b) = 素数

2818: Gcd

Description

Input

Output

Sample Input

Sample Output

HINT

4491. Primes in GCD Table

Problem code: PGCD

Input

Output

Example