HDU 5728 PowMod

最新推荐文章于 2020-03-17 17:06:06 发布

ShinFeb

最新推荐文章于 2020-03-17 17:06:06 发布

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分类专栏：欧拉函数

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$k=\sum_{i=1}^{m} \varphi (i*n)\ mod\ 1000000007$
$n$ 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 $k$

bblss123
BZOJ 3884 上帝集合的正确用法-PoPoQQQ

公式：

(1) $\sum_{i=1}^m \varphi(i*n)=\varphi(p)*\sum_{i=1}^m \varphi(i*n/p) + \sum_{i=1}^{floor(m/p)}\varphi(i*n)$

(2) $a^b\ mod \ p = a^{\varphi(p)+b\ mod \ \varphi(p)},b>=\varphi(p)$

(3) $a^{\varphi(n)}mod\ n=1,(a,n)=1$

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cassert>
#include<ctime>
#include<bitset>
#include<queue>
#include<set>
#define inf (1<<30)
#define INF (1ll<<62)
#define prt(x) cout<<#x<<":"<<x<<" "
#define prtn(x) cout<<#x<<":"<<x<<endl
using namespace std;
typedef long long ll;
template<class T>void sc(T &x){
    x=0;char c;int f=1;
    while(c=getchar(),c<48)if(c=='-')f=-1;
    do x=x*10+(c^48);
    while(c=getchar(),c>47);
    x*=f;
}
template<class T>void nt(T x){
    if(!x)return;
    nt(x/10);
    putchar('0'+x%10);
}
template<class T>void pt(T x){
    if(x<0)putchar('-'),x=-x;
    if(!x)putchar('0');
    else nt(x);
}
const int maxn=10000005;
const int mod=1e9+7;
int phi[maxn],sum[maxn],ai[maxn];
int prime[maxn/3],tot;
void init(){
    int limit=10000000;
    for(int i=2;i<=limit;i++){
        if(!phi[i])phi[i]=i-1,prime[tot++]=i,ai[i]=i;
        for(int k,j=0;j<tot;j++){
            if((k=i*prime[j])>limit)break;
            if(i%prime[j]==0){
                phi[k]=phi[i]*prime[j];
                ai[k]=prime[j];
                break;
            }
            phi[k]=phi[i]*(prime[j]-1);
            ai[k]=prime[j];
        }
    }
    phi[1]=1;ai[1]=1;
    for(int i=1;i<=limit;i++)
        sum[i]=(sum[i-1]+phi[i])%mod;
}
int n,m,p,K;
int calc(int n,int m){
    if(n==1)return sum[m];
    if(!m)return 0;//
    return (1ll*phi[ai[n]]*calc(n/ai[n],m)+calc(n,m/ai[n]))%mod;
}
ll qpow(ll a,ll b,int mod){
    ll c=1;
    for(;b;b>>=1,a=(a*a)%mod)
        if(b&1)c=(a*c)%mod;
    return c;
}
int pow(int p){
    if(p<=1)return 0;
    int q=pow(phi[p])+phi[p];
    return qpow(K,q,p);
}
void solve(){
    K=calc(n,m);
    int ans=pow(p);
    pt(ans);
    putchar('\n');
}
int main(){
//  freopen("pro.in","r",stdin);
//  freopen("chk.out","w",stdout);
    init();
    while(scanf("%d%d%d",&n,&m,&p)!=EOF)solve();
    return 0;
}