Problem Description
Goffi is doing his math homework and he finds an equality on his text book: gcd(n−a,n)×gcd(n−b,n)=nk.
Goffi wants to know the number of (a,b) satisfy the equality, if n and k are given and 1≤a,b≤n.
Note: gcd(a,b) means greatest common divisor of a and b.
Input
Input contains multiple test cases (less than 100). For each test case, there‘s one line containing two integers n and k (1≤n,k≤109).
Output
For each test case, output a single integer indicating the number of (a,b) modulo 109+7.
Sample Input
2 1 3 2
Sample Output
2 1
Hint
For the first case, (2, 1) and (1, 2) satisfy the equality.
Source
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 }