Beads of N colors are connected together into a circular necklace of N beads (N<=1000000000). Your job is to calculate how many different kinds of the necklace can be produced. You should know that the necklace might not use up all the N colors, and the repetitions that are produced by rotation around the center of the circular necklace are all neglected.
You only need to output the answer module a given number P.
Input
You only need to output the answer module a given number P.
The first line of the input is an integer X (X <= 3500)
representing the number of test cases. The following X lines each
contains two numbers N and P (1 <= N <= 1000000000, 1 <= P
<= 30000), representing a test case.
Output
For each test case, output one line containing the answer.
Sample Input5
1 30000
2 30000
3 30000
4 30000
5 30000
Sample Output
1
3
11
70
629
这题不用考虑翻转操作
所以答案就是$ans=sum_{i=1}^nn^{gcd(i,n)}$
枚举gcd，可以得到：
$ans=frac{1}{n}sum_{d=1}^nsum_{i=1}^{n}n^{d}[gcd(i,n)==d]$
$ans=sum_{d=1}^nn^{d-1}sum_{i=1}^{frac{n}{d}}[gcd(i,n/d)==1]$
$ans=sum_{d=1}^nn^{d-1}phi(frac{n}{d})$
$ans=sum_{d=1}^nn^{d-1}sum_{i=1}^{frac{n}{d}}[gcd(i,n/d)==1]$
$ans=sum_{d=1}^nn^{d-1}phi(frac{n}{d})$
外层枚举n因数
求欧拉函数用定义式可以O(√n)求解
这样常数很大
实现筛出素数，求欧拉时只枚举素数
因为n的因数远远小于√n，所以很快，虽然理论上会超
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 using namespace std; 7 typedef long long lol; 8 int tot,vis[50001],prime[50001],ans,Mod; 9 void pre() 10 {int i,j; 11 for (i=2;i<=50000;i++) 12 { 13 if (vis[i]==0) 14 { 15 tot++; 16 prime[tot]=i; 17 } 18 for (j=1;j<=tot;j++) 19 { 20 if (1ll*i*prime[j]>50000) break; 21 vis[i*prime[j]]=1; 22 if (i%prime[j]==0) 23 break; 24 } 25 } 26 } 27 int qpow(int x,int y) 28 { 29 int res=1; 30 while (y) 31 { 32 if (y&1) res=1ll*res*x%Mod; 33 x=1ll*x*x%Mod; 34 y>>=1; 35 } 36 return res; 37 } 38 int phi(int n) 39 {int i; 40 int s=n; 41 for (i=1;i<=tot&&1ll*prime[i]*prime[i]<=n;i++) 42 if (n%prime[i]==0) 43 { 44 s=s-s/prime[i]; 45 while (n%prime[i]==0) n/=prime[i]; 46 } 47 if (n!=1) 48 s=s-s/n; 49 return s%Mod; 50 } 51 int main() 52 {int T,tim,i,n; 53 cin>>T; 54 pre(); 55 while (T--) 56 { 57 scanf("%d%d",&n,&Mod); 58 tim=sqrt(n); 59 ans=0; 60 for (i=1;i<=tim;i++) 61 if (n%i==0) 62 { 63 ans=(ans+1ll*qpow(n,i-1)*phi(n/i)%Mod)%Mod; 64 if (i*i<n) 65 ans=(ans+1ll*qpow(n,n/i-1)*phi(i)%Mod)%Mod; 66 } 67 printf("%d ",ans); 68 } 69 }