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- 自己天真以为无法互质就没有办法降幂,其实还有广义欧拉定理!QAQ
- a^b mod c = a^(b mod phi(c) + phi(c))
- 相关知识及其证明:求幂大法(广义欧拉定理)及其证明
AC代码:
#include<bits/stdc++.h>
#define IO ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define pb(x) push_back(x)
#define sz(x) (int)(x).size()
#define sc(x) scanf("%d",&x)
#define pr(x) printf("%d\n",x)
#define abs(x) ((x)<0 ? -(x) : x)
#define all(x) x.begin(),x.end()
#define mk(x,y) make_pair(x,y)
#define debug printf("!!!!!!\n")
#define fin freopen("in.txt","r",stdin)
#define fout freopen("out.txt","w",stdout)
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
const int mod = 1e9+7;
const double PI = 4*atan(1.0);
const int maxm = 1e8+5;
const int maxn = 1e4+5;
const int INF = 0x3f3f3f3f;
const ll LINF = 1ll<<62;
ll phi(ll x)
{
ll ans = x;
for(int i=2;i*i<=x;i++)
{
if(x%i == 0){
ans = ans/i*(i-1);
while(x%i == 0) x /= i;
}
}
if(x>1) ans = ans/x*(x-1);
return ans;
}
ll qpow(ll a,ll b,ll m)
{
ll res = 1;
while(b){
if(1&b) res = (res*a)%m;
a = (a*a)%m;
b >>= 1;
}
return res;
}
ll solve(ll x,ll m) // a^b mod c = a^(b mod phi(c) + phi(c))
{
if(m == 1) return 0;
if(x == 1) return 1;
if(x == 2) return 2%m;
if(x == 3) return 9%m;
if(x == 4) return qpow(4,9,m);
else{
ll e = phi(m);
ll k = solve(x-1,e);
ll ans = qpow(x,k+e,m);
return ans;
}
}
int main()
{
// fin;
IO;
ll n,m;
cin>>n>>m;
cout<<solve(n,m)<<endl;
return 0;
}