欧拉降幂用来求解 a^b % mod问题
求解方法 :
a^b%mod
if(b > mod) return a ^ (b % phi[mod]+ phi[mod]) % mod;
else return a ^ (b % phi[mod]) % mod;
//phi(欧拉函数,小于n的与其互质的个数)
所以说要用到欧拉函数和快速幂。
快速幂
//快速幂可以调用下面的取模函数,实现上面的判断语句
inline ll Mod(ll x, ll mod){
return x < mod ? x : x % mod + mod;
}
inline ll Mod(ll x, ll mod){
return x < mod ? x : x % mod + mod;
}
ll _pow(ll a, ll b, ll c){
ll ans = 1;
while(b){
if(b & 1) ans = Mod(ans*a, c);
a = Mod(a*a, c);
b >>= 1;
}
return ans;
}
欧拉函数与欧拉筛:
const int N = 1e5 + 5;
int euler[N];
int Euler(int n){
int ans=n;
for(int i=2;i*i<=n;i++){
if(n%i==0){
ans=ans/i*(i-1);
while(n%i==0)n/=i;
}
}
if(n>1)ans=ans/n*(n-1);
return ans;
}
void get_euler(){
fill(euler,euler+N,0);
euler[1]=1;
for(int i=2;i<N;i++){
if(!euler[i]){
for(int j=i;j<N;j+=i){
if(!euler[j])euler[j]=j;
euler[j]=euler[j]/i*(i-1);
}
}
}
}
模板题1
求a^b % m,b最大是10^2e7,需要字符串输入
#include<bits/stdc++.h>
using namespace std;
#define debug(x) cout<<#x<<"="<<x<<endl
#define ll long long
#define lowbit(x) x&(-x)
#define ls p<<1
#define rs p<<1|1
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
inline int read(){
int p=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){p=(p<<1)+(p<<3)+(c^48),c=getchar();}
return f*p;
}
ll phi[maxn];
void get_euler(int N){
fill(phi,phi+N,0);
phi[1]=1;
for(int i=2;i<N;i++){
if(!phi[i]){
for(int j=i;j<N;j+=i){
if(!phi[j])phi[j]=j;
phi[j]=phi[j]/i*(i-1);
}
}
}
}
ll euler(ll n){
ll ans = n;
for(int i = 2; i * i <= n; i ++){
if(n % i == 0) {
ans = ans / i * (i - 1);
while(n % i == 0) n /= i;
}
}
if(n > 1) ans = ans / n * (n - 1);
return ans;
}
inline ll Mod(ll x, ll mod){
return x < mod ? x : x % mod + mod;
}
ll _pow(ll a, ll b, ll c){
ll ans = 1;
while(b){
if(b & 1) ans = Mod(ans*a, c);
a = Mod(a*a, c);
b >>= 1;
}
return ans;
}
ll read(ll m){
char ch = getchar();
while(!isdigit(ch)) ch = getchar();
int f = 0;
ll b = 0;
while(isdigit(ch)){
b = b * 10 + ch - '0';
b = Mod(b, m);
ch = getchar();
}
return b;
}
int main(){
ll a, m;
scanf("%lld %lld", &a, &m);
ll _b = read(euler(m));
printf("%lld\n", _pow(a, _b, m)%m);
return 0;
}
模板题2
求a^a^a^a...^a( b个a的幂次)%m,用dfs求解
#include<bits/stdc++.h>
using namespace std;
#define debug(x) cout<<#x<<"="<<x<<endl
#define lowbit(x) x&(-x)
#define ll long long
#define rs p<<1|1
#define ls p<<1
const int maxn = 1e6 + 5;
const int mod = 1e9 + 7;
inline ll read(){
ll p=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){p=(p<<1)+(p<<3)+(c^48),c=getchar();}
return f*p;
}
inline ll Mod(ll x, ll mod){
return x < mod ? x : x % mod + mod;
}
ll _pow(ll a, ll b, ll c){
ll ans = 1;
while(b){
if(b & 1) ans = Mod(ans*a, c);
a = Mod(a*a, c);
b >>= 1;
}
return ans;
}
ll euler[maxn];
void get_euler(int N){
fill(euler,euler+N,0);
euler[1]=1;
for(int i=2;i<N;i++){
if(!euler[i]){
for(int j=i;j<N;j+=i){
if(!euler[j])euler[j]=j;
euler[j]=euler[j]/i*(i-1);
}
}
}
}
ll dfs(ll a, ll cnt, ll c){
if(cnt == 0) return 1; //个数用完,返回1
if(c == 1) return a; //模是1,那么指数是b%1+1=1,那么a的1次,返回a
return _pow(a, dfs(a, cnt - 1, euler[c]), c);
}
void solve(){
get_euler(maxn);
ll t, a, cnt, c;
cin >> t;
while(t --){
cin >> a >> cnt >> c;
cout << dfs(a, cnt, c) % c << endl;
}
}
int main(){
solve();
return 0;
}