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hdu 4983 Goffi and GCD（欧拉函数）

Problem Description

Goffi is doing his math homework and he finds an equality on his text book:

Input

Input contains multiple test cases (less than 100). For each test case, there's one line containing two integers

Output

For each test case, output a single integer indicating the number of (

Sample Input

2 1 
3 2

Sample Output

2
1

Hint

For the first case, (2, 1) and (1, 2) satisfy the equality.

Source

BestCoder Round #6

发现自己欧拉函数都给忘记了，所有赶紧补题。。。

1、k!=1时情况很简单，记住将if(k==2 || n==1)这个特判放在if(k>2)的前面，因为这个WA了很久，各种原因自己思考。

2、下面讨论k=1时情况。x=gcd(n-a,n),则n/x=gcd(n-b,n),因为n-a可以取到0...n-1也就是1....n，所以完全可以去掉n-这个限制条件，即gcd(a,n)=x、gcd(b,n)=n/x时个数，因为a<=n，所以gcd(a,n)的个数=u[n/x],u是欧拉函数。所以原式等于sigma(u[n/x]*u[x])其中x是n的约数。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 #define MOD 1000000007
 6 #define ll long long
 7 ll eular(ll n)
 8 {
 9     ll res=1;
10     for(ll i=2;i*i<=n;i++)
11     {
12         if(n%i==0)
13         {
14             n/=i,res*=i-1;
15             while(n%i==0)
16             {
17                 n/=i;
18                 res*=i;
19             }
20         }
21     }
22     if(n>1) res*=n-1;
23     return res;
24 }
25 ll n,k;
26 int main()
27 {
28     while(scanf("%I64d%I64d",&n,&k)==2)
29     {
30         if(k==2 || n==1)
31         {
32             printf("1\n");
33             continue;
34         }
35         if(k>2)
36         {
37             printf("0\n");
38             continue;
39         }
40         
41         ll ans=0;
42         for(ll i=1;i*i<=n;i++)
43         {
44             if(n%i==0)
45             {
46                 if(i*i!=n)
47                   ans=(ans+eular(n/i)*eular(i)*2)%MOD;
48                 else 
49                     ans=(ans+eular(n/i)*eular(i))%MOD;
50             }
51         }
52         printf("%I64d\n",ans);
53 
54     }
55     return 0;
56 }