hdu 5728 （公式推导+指数循环节）
PowMod
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1221 Accepted Submission(s): 428
Problem Description
Declare:
k=∑（m，i=1）φ(i∗n) mod 1000000007
n is a square-free number.
φ is the Euler’s totient function.
find:
ans=k^(k^(k^(k…k))) mod p
There are infinite number of k
Input
Multiple test cases(test cases ≤100), one line per case.
Each line contains three integers, n,m and p.
1≤n,m,p≤107
Output
For each case, output a single line with one integer, ans.
Sample Input
1 2 6
1 100 9
Sample Output
4
7
这题比较难求的是k的值，只要知道k的值这题就是一个指数循环节定理的裸题了。
观察这个式子显然直接求k是不行的，发现n的因子中是没有重复的数的，这意味着当p整除n且i%p！=0时 phi(n/p*i*p)可以写成phi(p)*phi(i*n/p) 而当i%p==0时，也就是说i含有质因子p这个式子可以写成
phi(i*n)=phi((i* (n/p)*p)=p*phi((i/p)*n) 令i/p =j 所以可以写成p*phi(j*n)
有了这两个式子，则原式= ∑（1 <= i <= m && i%p!=0）phi(p)φ(i∗n/p) + ∑（1 <= j <= m /p && j%p ==0）p*phi(j*n) 发现p=phi(p)+1 于是两个式子能和并在一起即 ∑（1 <= i <= m ）phi(p)φ(i∗n/p) +∑（1 <= i <= m / p ）φ(i∗n) 就可以递归的算出k的值了，后面一部分则是应用指数循环节定理A^B%C = A^(B%phi(C)+phi(C));所以依旧可以递归的求出。
#include if(!isp[i])
{
p[isp[0]++]=i;
phi[i]=i-1;
}
for(int j=0;j0] && i*p[j]true;
if(i%p[j]==0)
{
phi[i*p[j]]=phi[i]*p[j];
break;
}
else phi[i*p[j]]=phi[i]*(p[j]-1);
}
}
sum[0]=0;
for(int i=1;i1]+phi[i])%mod;
}
long long fk(long long m,long long n,int deep)
{
if (m==0) return 0;
if (n==1) return sum[m];
return (phi[rec[deep]]*fk(m,n/rec[deep],deep+1)%mod+fk(m/rec[deep],n,deep))%mod;
}
long long f(long long x,long long k,long long md)
{
long long res=1;
while(k){
if(k&1) res=res*x%md;
k>>=1;
x=x*x%md;
}
return res;
}
long long solve(long long k,long long md)
{
if (md==1) return 1;
return f((long long)k,solve(k,phi[md])+phi[md],md);
}
int main(){
init();
while (scanf("%I64d%I64d%I64d",&n,&m,&pp)!=EOF)
{
cnt=0;
long long tmp=n;
for (int i=0;i0];i++)
{
if (tmp%p[i]==0)
{
rec[cnt++]=p[i];
tmp/=p[i];
}
if (tmp==1) break;
}
if (tmp!=1) rec[cnt++]=tmp;
long long k=fk(m,n,0)%mod;
printf("%I64d\n",solve(k,pp)%pp);
}
}