大数欧拉函数 + 降幂 —

本文主要是介绍大数欧拉函数 + 降幂 —— 模板题，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题目链接：点我啊！！~~~

$P S ：$ 注意要判断扩展欧拉降幂，并用快速乘

$O (s q r t (n))$ 求欧拉函数:

#include<bits/stdc++.h>
#define rint register int
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
using namespace std;
typedef long long ll;
typedef pair <int,int> pii;
const int maxn = 2e7 + 5;
ll a, p;
char b[maxn];ll qmul(ll a, ll b) {return (a*b - (ll)((long double)a/p*b)*p+p)%p;
}ll qpow(ll a, ll b){ll ret = 1;while(b){if(b & 1) ret = qmul(ret, a);a = qmul(a, a);b >>= 1;}return ret;
}ll getphi(ll x){ll ret = x;for(ll i=2; i*i<=x; i++)if(x%i==0){ret -= ret / i;while(x%i==0) x /= i;}if(x > 1) ret -= ret / x;return ret;
}int main() {scanf("%lld%lld%s", &a, &p, &b);ll ans = 0, phi = getphi(p), f = 0;for(int i=0; b[i]; i++){ans = ans * 10 + b[i] - '0';if(ans >= phi) f = 1;ans %= phi;}if(f) ans = qpow(a, ans+phi);else ans = qpow(a, ans);printf("%lld\n", ans);
}

利用 $M i l l e r - R a b i n$ 求质因子后求得:

#include<bits/stdc++.h>
#define rint register int
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
using namespace std;
typedef long long ll;
typedef pair <int,int> pii;
const int maxn = 2e7 + 5;namespace Factor {
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define PLL pair<ll,ll>
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)const int N=1010000;ll C,fac[10010],n,mut,a[1001000];int T,cnt,i,l,prime[N],p[N],psize,_cnt;ll _e[100],_pr[100];vector<ll> d;inline ll mul(ll a,ll b,ll p) {if (p<=1000000000) return a*b%p;else if (p<=1000000000000ll) return (((a*(b>>20)%p)<<20)+(a*(b&((1<<20)-1))))%p;else {ll d=(ll)floor(a*(long double)b/p+0.5);ll ret=(a*b-d*p)%p;if (ret<0) ret+=p;return ret;}}void prime_table() {int i,j,tot,t1;for (i=1; i<=psize; i++) p[i]=i;for (i=2,tot=0; i<=psize; i++) {if (p[i]==i) prime[++tot]=i;for (j=1; j<=tot && (t1=prime[j]*i)<=psize; j++) {p[t1]=prime[j];if (i%prime[j]==0) break;}}}void init(int ps) {psize=ps;prime_table();}ll powl(ll a,ll n,ll p) {ll ans=1;for (; n; n>>=1) {if (n&1) ans=mul(ans,a,p);a=mul(a,a,p);}return ans;}bool witness(ll a,ll n) {int t=0;ll u=n-1;for (; ~u&1; u>>=1) t++;ll x=powl(a,u,n),_x=0;for (; t; t--) {_x=mul(x,x,n);if (_x==1 && x!=1 && x!=n-1) return 1;x=_x;}return _x!=1;}bool miller(ll n) {if (n<2) return 0;if (n<=psize) return p[n]==n;if (~n&1) return 0;for (int j=0; j<=7; j++) if (witness(rand()%(n-1)+1,n)) return 0;return 1;}ll gcd(ll a,ll b) {ll ret=1;while (a!=0) {if ((~a&1) && (~b&1)) ret<<=1,a>>=1,b>>=1;else if (~a&1) a>>=1;else if (~b&1) b>>=1;else {if (a<b) swap(a,b);a-=b;}}return ret*b;}ll rho(ll n) {for (;;) {ll X=rand()%n,Y,Z,T=1,*lY=a,*lX=lY;int tmp=20;C=rand()%10+3;X=mul(X,X,n)+C;*(lY++)=X;lX++;Y=mul(X,X,n)+C;*(lY++)=Y;for(; X!=Y;) {ll t=X-Y+n;Z=mul(T,t,n);if(Z==0) return gcd(T,n);tmp--;if (tmp==0) {tmp=20;Z=gcd(Z,n);if (Z!=1 && Z!=n) return Z;}T=Z;Y=*(lY++)=mul(Y,Y,n)+C;Y=*(lY++)=mul(Y,Y,n)+C;X=*(lX++);}}}void _factor(ll n) {for (int i=0; i<cnt; i++) {if (n%fac[i]==0) n/=fac[i],fac[cnt++]=fac[i];}if (n<=psize) {for (; n!=1; n/=p[n]) fac[cnt++]=p[n];return;}if (miller(n)) fac[cnt++]=n;else {ll x=rho(n);_factor(x);_factor(n/x);}}void dfs(ll x,int dep) {if (dep==_cnt) d.pb(x);else {dfs(x,dep+1);for (int i=1; i<=_e[dep]; i++) dfs(x*=_pr[dep],dep+1);}}void norm() {sort(fac,fac+cnt);_cnt=0;rep(i,0,cnt) if (i==0||fac[i]!=fac[i-1]) _pr[_cnt]=fac[i],_e[_cnt++]=1;else _e[_cnt-1]++;}vector<ll> getd() {d.clear();dfs(1,0);return d;}vector<ll> factor(ll n) {cnt=0;_factor(n);norm();return getd();}vector<PLL> factorG(ll n) {cnt=0;_factor(n);norm();vector<PLL> d;rep(i,0,_cnt) d.pb(mp(_pr[i],_e[i]));return d;}ll getphi(ll x) { // 2019/9/17vector <PLL> G;G = factorG(x);double ret = x;rep(i,0,G.size())ret *= (1.0 - 1.0/(double)G[i].first);return ret;}bool is_primitive(ll a,ll p) {assert(miller(p));vector<PLL> D=factorG(p-1);rep(i,0,SZ(D)) if (powl(a,(p-1)/D[i].fi,p)==1) return 0;return 1;}
#undef fi
#undef pb
#undef se
#undef mp
#undef PLL
#undef SZ
#undef all
#undef rep
#undef per
};ll qmul(ll a, ll b, ll p) {return (a*b - (ll)((long double)a/p*b)*p+p)%p;
}ll qpow(ll a, ll b, ll p){ll ret = 1;while(b){if(b & 1) ret = qmul(ret, a, p);a = qmul(a, a, p);b >>= 1;}return ret;
}char b[maxn];
int main() {ll a, p;Factor::init(1e4 + 5);scanf("%lld%lld%s", &a, &p, &b);ll ans = 0, phi = Factor::getphi(p), f = 0;for(int i=0; b[i]; i++){ans = ans * 10 + b[i] - '0';if(ans >= phi) f = 1;ans %= phi;}if(f) ans = qpow(a, ans+phi, p);else ans = qpow(a, ans, p);printf("%lld\n", ans);
}

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