2019 南京ICPC网络赛 B. super_log

题目链接： https://nanti.jisuanke.com/t/41299

题意：计算迭代次幂 的值。

扩展欧拉定理

先使用Eratosthenes筛或者线性筛求出1~N每个数的欧拉函数值，然后通过递归求解。

方法1：保守写法

# include <iostream>

const int N = 1e6 + 5;

typedef long long ll;
ll phi[N];

void init(int n) {
   
	phi[1] = 1;
	for (int i = 2; i <= n; i++) {
   
		phi[i] = i;
	}

	for (int i = 2; i <= n; i++) {
   
		if (phi[i] == i) {
   
			for (int j = i; j <= n; j += i) {
   
				phi[j] = phi[j] / i * (i - 1);
			}
		}
	}
}

ll qpow(ll x, ll p, ll mod) {
   
	ll ans = 1;

	while (p) {
   
		if (p & 1) {
   
			ans = ans * x % mod;
		}

		x = x * x % mod;
		p >>= 1;
	}

	return ans;
}

ll solve(ll a, ll b, ll p) {
   
	if (b == 0) {
   
		return 1;	
	}

	if (p == 1) {
   
		return 0;		// AC
// return 1; // AC
	}

	ll P = solve(a, b - 1, phi[p]);

	if (P < phi[p] && P) {
   
		return qpow(a, P, p);
	} else {
   
		return qpow(a, P + phi[p], p);
	}
}

int main() {
   
	init(N - 5);

	int T;
	std::cin >> T;

	while (T--) {
   
		ll a;
		ll b;
		ll mod;
		std::cin >> a >> b >> mod;

		ll ans = solve(a, b, mod);

		std::cout << (ans % mod) << "\n";
	}

	return 0;
}

方法2：在快速幂取模的时候把需要取模的数加上一个模数，递归时直接一次解出。

# include <iostream>
# include <cstdio>

const int N = 1e6 + 5;

typedef long long ll;

ll phi[N];    

void init(int n) {
   
	phi[1] = 1;
	for (int i = 2; i <= n; i++) {
   
		phi[i] = i;
	}

	for (int i = 2; i <= n; i++) {
   
		if (phi[i] == i) {
   
			for (int j = i; j <= n; j += i) {
   
				phi[j] = phi[j] / i * (i - 1);
			}
		}
	}
}
 
ll Mod(ll a, ll b) {
   
	return a < b ? a : (a % b + b);
}
 
ll qpow(ll x, ll p, ll mod) {
   
	ll ans = 1 % mod;
	while (p) {
   
		if (p & 1)
			ans = Mod(ans * x, mod);
			
		x = Mod(x * x, mod);
		p >>= 1;
	}
	 
	return ans;
}

ll solve(ll a, ll b, ll p) {
   
	if (b == 0) {
   
		return 1;
	}
	
	if (p == 1) {
   
		return 1;		// AC
// return 0; // WA
	}
	
	return qpow(a, solve(a, b - 1, phi[p]), p);
}

int main() {
   	
	init(N - 5);
	
	int T;
	scanf("%d", &T);
	
	while (T--) {
   
		ll a;
		ll b;
		ll mod;
		scanf("%lld %lld %lld", &a, &b, &mod);
				
		printf("%lld\n", solve(a, b, mod) % mod);
	}
	
	return 0;
}

那个注释部分，p==1时的返回值还没有弄懂