Description
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.
现在有N种颜色的珠子，我们要做一个N个珠子组成的环形项链（N≤1000000000）。
你的任务是计算可以生产多少种不同的项链。注意项链并不一定要用到所有颜色的珠子，而且旋转之后等同的项链被认为是一样的。
只需要输出这个总数Mod P的余数。
Input
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 Input
5
1 30000
2 30000
3 30000
4 30000
5 30000
Sample Output
1
3
11
70
629
polya+数论
考虑到polya朴素式子(sumlimits_{i=1}^{n} c^{gcd(i,n)})(c为颜色数)，n到了1e9之后必定超时……
我们换个想法，考虑枚举n的每个因数d，那么(gcd(n,i)=d)的数共有多少个呢？共(phi(frac{n}{i}))个
为什么是这么多呢？
设(gcd(x,n)=i)，则有(gcd(frac{x}{i},frac{n}{i})=1)，所以说有多少个小于等于n的x满足(gcd(frac{x}{i},frac{n}{i})=1)，即满足(gcd(x,n)=i)，因此，它们的个数为(phi(frac{n}{i}))个
有了这个，我们就可以得到$$Ans=sum_{d|n} n^d*phi(frac{n}{d})$$
最后总状态还要除上一个n，还要对p取模，这个很简单，写个高精度。每个(sum)提个n出来就好了啊
/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline int read(){
int x=0,f=1;char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x>=10) print(x/10);
putchar(x%10+'0');
}
const int N=1e5;
int prime[N+10];
bool inprime[N+10];
int tot,n,p,Ans;
void prepare(){
for (int i=2;i<=N;i++){
if (!inprime[i]) prime[++tot]=i;
for (int j=1;j<=tot&&i*prime[j]<=N;j++){
inprime[i*prime[j]]=1;
if (i%prime[j]==0) break;
}
}
}
int phi(int x){
int ans=x;
for (int i=1;prime[i]*prime[i]<=x;i++){
if (x%prime[i]) continue;
ans-=ans/prime[i];
while (x%prime[i]==0) x/=prime[i];
}
if (x!=1) ans-=ans/x;
return ans;
}
int mlt(int a,int b){
int res=1;
for (;b;b>>=1,a=1ll*a*a%p) if (b&1) res=1ll*res*a%p;
return res;
}
int main(){
prepare();
for (int Data=read();Data;Data--){
n=read(),p=read(),Ans=0;
for (int i=1;i*i<=n;i++){
if (n%i) continue;
int j=n/i;
Ans=(Ans+1ll*mlt(n,i-1)*phi(j))%p;
if (i!=j) Ans=(Ans+1ll*mlt(n,j-1)*phi(i))%p;
//省时间，枚举到sqrt(n)即可，判下是否为完全平方数
}
printf("%d
",Ans);
}
return 0;
}