[MtOI2019]幽灵军团
题意
求下面表达式的值，
\[\prod_{i=1}^{A} \prod_{j=1}^{B} \prod_{k=1}^{C} \left( \frac{lcm(i,j)}{gcd(i,k)} \right)^{f(type)} \pmod{p}
\]
其中，\(type=\{0,1,2\}\)，\(f(type)\)满足：
\[f(0)=1,
\]
\[f(1)=i \times j \times k,
\]
\[f(2) = gcd(i,j,k).
\]
其中， \(A,B,C \leqslant 10^5.\)
Solution
type = 0
拿到这个式子非常兴奋，然后就开始疯狂化简，最终化简成了一个\(O(n\sqrt{n} \log {n})\)的表达式，非常沮丧的是不能满足题目的要求。
我们重新分析这个式子，
\[\prod_{i=1}^{A} \prod_{j=1}^{B} \prod_{k=1}^{C} \left( \frac{lcm(i,j)}{gcd(i,k)} \right)
\]
\[=\prod_{i=1}^{A} \prod_{j=1}^{B} \prod_{k=1}^{C} \left( \frac{i \times j}{gcd(i,j) \times gcd(i,k)} \right)
\]
不难发现，这个式子的\(i,j,k\)是对称的，相当于我们仅需要计算以下两部分式子的值，然后将其组合起来，
\[\prod_{i=1}^{A} \prod_{j=1}^{B} \prod_{k=1}^{C} i
\]
\[\prod_{i=1}^{A} \prod_{j=1}^{B} \prod_{k=1}^{C} \frac{1}{gcd(i,j)}
\]
分别来对两个式子进行化简
\[\prod_{i=1}^{A} \prod_{j=1}^{B} \prod_{k=1}^{C} i
\]
\[= \prod_{i=1}^{A} i ^ {BC}
\]
\[= \left( \prod_{i=1}^{A} i \right) ^ {BC}
\]
可以通过预处理前缀积\(O(\log{n})\)实现对于此式的计算。
\[\prod_{i=1}^{A} \prod_{j=1}^{B} \prod_{k=1}^{C} \frac{1}{gcd(i,j)} \pmod{p}
\]
\[=\left(\prod_{i=1}^{A} \prod_{j=1}^{B} \prod_{k=1}^{C} gcd(i,j) \right)^{-1} \pmod{p}
\]
接下来只计算其本身的值，然后再计算逆元求出答案。
\[\prod_{i=1}^{A} \prod_{j=1}^{B} \prod_{k=1}^{C} gcd(i,j)
\]
\[=\left(\prod_{i=1}^{A} \prod_{j=1}^{B} gcd(i,j)\right)^{C}
\]
\[\Large {=\left(\prod_{d=1}^{min\{A,B,C\}} d ^{\sum_{i=1}^{\lfloor \frac{A}{d} \rfloor} \sum_{j=1}^{\lfloor \frac{B}{d} \rfloor } \sum_{x|i, x|j} \mu(x)}\right)^{C}}
\]
跟[SDOI2017]数字表格类似的，则有，
\[\Large {\left(\prod_{T=1}^{min\{A,B\}} ( \prod_{d|T} d^{\mu(\frac{T}{x})} ) ^ {\lfloor \frac{A}{T} \rfloor \lfloor \frac{B}{T} \rfloor}\right) ^ C }
\]
里面的部分在上述题目中已经处理过了，只需要计算最终得到值的\(C\)次幂，便可以计算出此答案。
type = 1
大致的思路显然跟上面类似，重新推一遍两个式子即可。
\[\prod_{i=1}^{A} \prod_{j=1}^{B} \prod_{k=1}^{C} i ^ {ijk}
\]
\[=\prod_{i=1}^{A} \prod_{j=1}^{B} \prod_{k=1}^{C} (i ^ {i}) ^ {jk}
\]
\[= (\prod_{i=1}^{A} i ^ {i}) ^ {\sum_{j=1}^{B} \sum_{k=1}^{C} jk}
\]
记\(S(n) = \sum_{i=1}^{n} i = \frac{n \times (n + 1)}{2}\)，则有，
\[= (\prod_{i=1}^{A} i ^ {i}) ^ {S(B) \ S(C)}
\]
与\(type=0\)类似的，我们可以通过预处理 \(i^i\)的前缀积，然后可以在\(O(\log{n})\)的时间复杂度内求解这一部分的值。
\[\prod_{i=1}^{A} \prod_{j=1}^{B} \prod_{k=1}^{C} (gcd(i,j))^{ijk}
\]
\[=\left(\prod_{i=1}^{A} \prod_{j=1}^{B} (gcd(i,j))^{ij} \right) ^ {S(C)}
\]
\[\Large{=\left(\prod_{d=1}^{min\{A,B\}} d^{d^2 \sum_{i=1}^{\lfloor \frac{A}{d} \rfloor} \sum_{j=1}^{\lfloor \frac{B}{d} \rfloor} i \times j \sum_{x|i, x|j}\mu(x)} \right) ^ {S(C)}}
\]
\[\Large{=\left(\prod_{d=1}^{min\{A,B\}} d^{d^2 \sum_{x=1}^{min\{\lfloor \frac{A}{d} \rfloor,\lfloor \frac{B}{d} \rfloor\}}x^2\mu(x)S(\lfloor \frac{A}{xd} \rfloor)\ S(\lfloor \frac{B}{xd} \rfloor)} \right) ^ {S(C)}}
\]
记\(T=xd\)，则有，
\[\large{\left( \prod_{T=1}^{min\{A,B\}} ( \prod_{d|T} d^{d^2 \mu(\frac{T}{d})} )^{S(\lfloor \frac{A}{T} \rfloor) S(\lfloor \frac{B}{T} \rfloor)} \right) ^ {S(C)} }
\]
类似\(type=0\)，不难发现 \(\prod_{d|T} d ^ {d^2 \mu(\frac{T}{d})}\) 可以枚举因子进行计算，在 \(O(n\sqrt{n})\) 的时间复杂度下完成预处理。
然后对\(T\)进行数论分块，可以在 \(O(\log{n} \sqrt{n}\) 下完成整个式子的计算。
type = 2
ln大法好
特别感谢洛谷题解区peterwuyihong的idea。
对于比较复杂的指数问题，我们可以想到将其进行\(ln\)和\(exp\)操作。
\[\prod_{i=1}^A \prod_{j=1}^B \prod_{k=1}^{C} \left( \frac{lcm(i,j)}{gcd(i,k) }\right) ^ {gcd(i,j,k)}
\]
对他取\(ln\)操作：
\[\sum_{i=1}^A \sum_{j=1}^B \sum_{k=1}^{C} \ln{\left( \frac{lcm(i,j)}{gcd(i,k) }\right)} \times gcd(i,j,k)
\]
莫比乌斯经典枚举因子，则有，
\[= \sum_{d=1}^{min\{A,B,C\}} \sum_{i=1}^A \sum_{j=1}^B \sum_{k=1}^C \ln{ ( \frac{lcm(i,j)}{gcd(i,k)} ) } \times d \times [gcd(i,j,k)=d]
\]
\[= \sum_{d=1}^{min\{A,B,C\}} d \sum_{i=1}^{\lfloor \frac{A}{d} \rfloor} \sum_{j=1}^{\lfloor \frac{B}{d} \rfloor} \sum_{k=1}^{\lfloor \frac{C}{d} \rfloor} \ln{ ( \frac{lcm(id,jd)}{gcd(id,kd)} ) } \times [gcd(i,j,k)=1]
\]
\[= \sum_{d=1}^{min\{A,B,C\}} d\sum_{i=1}^{\lfloor \frac{A}{d} \rfloor} \sum_{j=1}^{\lfloor \frac{B}{d} \rfloor} \sum_{k=1}^{\lfloor \frac{C}{d} \rfloor} \ln{ ( \frac{lcm(id,jd)}{gcd(id,kd)} ) } \sum_{x|gcd(i,j,k)} \mu(x)
\]
\[= \sum_{d=1}^{min\{A,B,C\}} d\sum_{x=1}^{min\{\lfloor \frac{A}{d} \rfloor,\lfloor \frac{B}{d} \rfloor\lfloor \frac{C}{d} \rfloor\}} \mu(x) \sum_{i=1}^{\lfloor \frac{A}{xd} \rfloor} \sum_{j=1}^{\lfloor \frac{B}{xd} \rfloor} \sum_{k=1}^{\lfloor \frac{C}{xd} \rfloor} \ln{ ( \frac{lcm(ixd,jxd)}{gcd(ixd,kxd)} ) }
\]
记\(T=xd\)，则有
\[= \sum_{T=1}^{min\{A,B,C\}} \sum_{d|T} (\mu(d) \frac{T}{d} \sum_{i=1}^{\lfloor \frac{A}{T} \rfloor} \sum_{j=1}^{\lfloor \frac{B}{T} \rfloor} \sum_{k=1}^{\lfloor \frac{C}{T} \rfloor} \ln{ ( \frac{lcm(iT,jT)}{gcd(iT,kT)} )}
\]
\[= \sum_{T=1}^{min\{A,B,C\}} \sum_{d|T} (\mu(d) \frac{T}{d} )\sum_{i=1}^{\lfloor \frac{A}{T} \rfloor} \sum_{j=1}^{\lfloor \frac{B}{T} \rfloor} \sum_{k=1}^{\lfloor \frac{C}{T} \rfloor} \ln{ ( \frac{lcm(i,j)}{gcd(i,k)} )}
\]
\[= \sum_{T=1}^{min\{A,B,C\}} \varphi(T)\sum_{i=1}^{\lfloor \frac{A}{T} \rfloor} \sum_{j=1}^{\lfloor \frac{B}{T} \rfloor} \sum_{k=1}^{\lfloor \frac{C}{T} \rfloor} \ln{ ( \frac{lcm(i,j)}{gcd(i,k)} )}
\]
然后将其\(exp\)一下，得到结果，
\[= \prod_{T=1}^{min\{A,B,C\}} (\prod_{i=1}^{\lfloor \frac{A}{T} \rfloor} \prod_{j=1}^{\lfloor \frac{B}{T} \rfloor} \prod_{k=1}^{\lfloor \frac{C}{T} \rfloor} \frac{lcm(i,j)}{gcd(i,k)} )^{\varphi(T)}
\]
\[= \prod_{T=1}^{min\{A,B,C\}} ANS_{type=0}(\lfloor \frac{A}{T} \rfloor,\lfloor \frac{B}{T} \rfloor,\lfloor \frac{C}{T} \rfloor)^{\varphi(T)}
\]
然后可以调用\(type=1\)的结果，就可以完成计算。
官方题解
与上面类似，我们分别对两个式子化简。
\[\prod_{i=1}^A \prod_{j=1}^B \prod_{k=1}^C i ^ {gcd(i,j,k)}
\]
\[\Large= {\prod_{d=1}^{min\{A,B,C\}} \prod_{i=1}^{\lfloor \frac{A}{d} \rfloor} (id) ^ {d\sum_{j=1}^{\lfloor \frac{B}{d} \rfloor} \sum_{k=1}^{\lfloor \frac{C}{d} \rfloor }[gcd(i,j,k)=1] } }
\]
\[\Large= {\prod_{d=1}^{min\{A,B,C\}} \prod_{i=1}^{\lfloor \frac{A}{d} \rfloor} (id) ^ {d\sum_{x=1}^{min\{\lfloor \frac{B}{d} \rfloor, \lfloor \frac{C}{d} \rfloor \} } \mu(x) \lfloor \frac{B}{xd} \rfloor \lfloor \frac{C}{xd} \rfloor } }
\]
\[\Large= {\prod_{d=1}^{min\{A,B,C\}} \prod_{x=1}^{min\{\lfloor \frac{A}{d} \rfloor, \lfloor \frac{B}{d} \rfloor, \lfloor \frac{C}{d} \rfloor \} } \prod_{i=1}^{\lfloor \frac{A}{xd} \rfloor} (ixd) ^ {d \ \mu(x) \lfloor \frac{B}{xd} \rfloor \lfloor \frac{C}{xd} \rfloor } }
\]
记\(T=xd\)，则有，
\[\Large= {\prod_{T=1}^{min\{A,B,C\}} \prod_{d | T} \prod_{i=1}^{\lfloor \frac{A}{T} \rfloor} (iT) ^ {d \ \mu(\frac{T}{d}) \lfloor \frac{B}{T} \rfloor \lfloor \frac{C}{T} \rfloor } }
\]
分离内层的\(i \times T\)，则有
\[\Large= {\prod_{T=1}^{min\{A,B,C\}} \left(\prod_{d | T} [ (\prod_{i=1}^{\lfloor \frac{A}{T} \rfloor} i) \times T^{\lfloor \frac{A}{T} \rfloor }]^ {d \ \mu(\frac{T}{d})} \right)^{ \lfloor \frac{B}{T} \rfloor \lfloor \frac{C}{T} \rfloor } }
\]
可以发现通过对内部函数\(O(n\sqrt{n}\log{n})\)的预处理，可以在\(O(\sqrt{n}\log{n})\)的时间复杂度内数论分块求解，但是先不着急求解，我们可以先计算下一个表达式。
\[\prod_{i=1}^A \prod_{j=1}^B \prod_{k=1}^C (gcd(i,j)^{gcd(i,j,k)})
\]
枚举情况更为复杂的\(gcd(i,j,k)\)，
\[\Large = {\prod_{d=1}^{min\{A,B,C\}} \prod_{i=1}^{\lfloor \frac{A}{d} \rfloor} \prod_{j=1}^{\lfloor \frac{B}{d} \rfloor } (gcd(id,jd))^{d \sum_{k=1}^{\lfloor \frac{C}{d} \rfloor} [gcd(i,j,k) = 1]} }
\]
（请允许我跳过一些步骤qwq）
\[\Large = {\prod_{d=1}^{min\{A,B,C\}} \prod_{x=1}^{min\{\lfloor \frac{A}{d} \rfloor,\lfloor \frac{B}{d} \rfloor,\lfloor \frac{C}{d} \rfloor\}}\prod_{i=1}^{\lfloor \frac{A}{xd} \rfloor} \prod_{j=1}^{\lfloor \frac{B}{xd} \rfloor } (xd \ gcd(i,j))^{d \ \mu(x)\lfloor \frac{C}{xd} \rfloor} }
\]
与上面类似，可以尝试分离\(xd\)和\(gcd(i,j)\)
对于\(xd\)分离出的部分，我们记\(T=xd\)，化简则有，
\[原式=\prod_{T=1}^{min\{A,B,C\}} \left(\prod_{d | T} [ T^{\lfloor \frac{A}{T} \rfloor }]^ {d \ \mu(\frac{T}{d})} \right)^{ \lfloor \frac{B}{T} \rfloor \lfloor \frac{C}{T} \rfloor }
\]
发现这一部分恰好为上一个式子所处理出的结果的一部分，又因为这一部分作为分母计入答案，所以说上下完全可以消掉，不再需要计算这一部分。
这个结论是将\(\prod \prod \prod i^{f(2)}\)和\(\prod \prod \prod gcd(i,j)^{f(2)}\)组合，但是实际中还有\(k\)的加入，其实可以通过将\(j\)和\(gcd(i,j)\)组合，将\(i\)与\(gcd(i,k)\)组合，就可以完美的使用两次这个结论。
接下来化简第二个式子中剩余的部分。
\[\Large {\prod_{d=1}^{min\{A,B,C\}} \prod_{x=1}^{min\{\lfloor \frac{A}{d} \rfloor,\lfloor \frac{B}{d} \rfloor,\lfloor \frac{C}{d} \rfloor\}}\prod_{i=1}^{\lfloor \frac{A}{xd} \rfloor} \prod_{j=1}^{\lfloor \frac{B}{xd} \rfloor } ( gcd(i,j))^{d \ \mu(x)\lfloor \frac{C}{xd} \rfloor} }
\]
取出其中的部分式子，
\[\prod_{i=1}^{\lfloor \frac{A}{xd} \rfloor} \prod_{j=1}^{\lfloor \frac{B}{xd} \rfloor } gcd(i,j)
\]
这个式子是经典的，化简过程略。
\[\Large{=\prod_{d'=1} \prod_{x'=1} d'^{\mu(x') \lfloor \frac{A}{xx'dd'} \rfloor \lfloor \frac{B}{xx'dd'} \rfloor}}
\]
代回原表达式，记\(T=xd, T'=x'd'\)，化简则有（上标略），
\[\Large{\prod_{T=1}} \left( \prod_{T'=1} \left( \prod_{d|T'}d^{\mu(\frac{T'}{d})} \right)^{\lfloor \frac{A}{TT'} \rfloor \lfloor \frac{B}{TT'} \rfloor} \right)^{\varphi(t) \lfloor \frac{C}{T} \rfloor}
\]
最里面的部分在\(type=0\)时已经处理过了，外层的两个循环使用两次数论分块即可，时间复杂度\(O(n^{\frac{3}{4}} \log{n}).\)
最后将第一个式子剩余部分乘上第二个式子剩余部分的逆元，即可求出答案的值。
卡常
在计算的过程中，会发现诸多预处理的变量中，有大量的需要求解其逆元，并且若干\(\sum_{i=1}^n i\)或者\(\prod_{i=1}^n i\)需要计算，可以通过预先处理其值，然后在后面的计算中直接调用，可以有效地提高程序的运行效率。
Code
点击查看代码
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
const int N = 1e6 + 50;
inline int min(register int a, register int b) {
return (a < b ? a : b);
}
inline int read() {
register int res = 0, f = 1;
register char c = getchar();
while (c > '9' || c < '0') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
res = (res << 3) + (res << 1) + c - '0';
c = getchar();
}
return res * f;
}
int mod;
inline int add_mod(register int a, register int b) {
a += b;
if (a < 0) a += mod;
if (a >= mod) a -= mod;
return a;
}
inline int del_mod(register int a, register int b) {
a -= b;
if (a < 0) a += mod;
if (a >= mod) a -= mod;
return a;
}
inline int mul_mod(register int a, register int b) {
a = 1ll * a * b % mod;
return a;
}
int num;
int prime[N], mu[N];
int v[N];
int mul[N], pmul[N];
int g[N], invg[N];
int F[N], inv[N], G[N];
int S[N];
int invF[N], invG[N];
int phi[N], sumphi[N];
inline int qpow(register int a, register int b) {
register int base = 1;
while (b) {
if (b & 1) base = mul_mod(base, a);
a = mul_mod(a, a);
b >>= 1;
}
return base;
}
inline void init() {
register int MAX = 1e5;
F[0] = F[1] = mu[1] = 1;
inv[0] = inv[1] = 1;
pmul[1] = mul[1] = 1;
g[1] = G[0] = G[1] = 1;
invg[0] = invg[1] = 1;
phi[1] = 1;
S[1] = 1;
for (register int i = 2; i <= MAX; ++i) {
S[i] = (1ll * i * (i + 1) / 2) % (mod - 1);
F[i] = G[i] = 1;
invF[0] = invF[1] = invG[0] = invG[1] = 1;
mul[i] = mul_mod(mul[i - 1], i);
pmul[i] = mul_mod(pmul[i - 1], qpow(i, i));
g[i] = i;
invg[i] = qpow(g[i], mod - 2);
inv[i] = qpow(i, mod - 2);
if (!v[i]) {
prime[++num] = i;
v[i] = i;
mu[i] = -1;
phi[i] = i - 1;
}
for (register int j = 1; j <= num; ++j) {
if (prime[j] > v[i] || prime[j] > MAX / i) break;
v[i * prime[j]] = prime[j];
if (i % prime[j] == 0) {
phi[i * prime[j]] = phi[i] * prime[j];
mu[i * prime[j]] = 0;
break;
}
phi[i * prime[j]] = phi[i] * phi[prime[j]];
mu[i * prime[j]] = -mu[i];
}
}
for (register int i = 1; i <= MAX; ++i) {
sumphi[i] = (sumphi[i - 1] + phi[i]) % (mod - 1);
if (!mu[i]) continue;
for (register int j = i; j <= MAX; j += i) {
F[j] = mul_mod(F[j], (mu[i] == 1 ? j / i : inv[j / i]));
G[j] = mul_mod(G[j], qpow((mu[i] == 1 ? g[j / i] : invg[j / i]), (1ll * j * j % (mod - 1))));
}
}
for (register int i = 2; i <= MAX; ++i) {
F[i] = mul_mod(F[i], F[i - 1]);
G[i] = mul_mod(G[i], G[i - 1]);
invF[i] = qpow(F[i], mod - 2);
invG[i] = qpow(G[i], mod - 2);
}
}
inline int calc1(register int A, register int B, register int C) {
register int res = qpow(mul[A], (1ll * B * C % (mod - 1)));
res = mul_mod(res, qpow(mul[B], (1ll * A * C % (mod - 1))));
register int tmp = 1;
for (register int i = 1, j; i <= min(A, B); i = j + 1) {
j = min(A / (A / i), B / (B / i));
tmp = mul_mod(tmp, qpow(mul_mod(F[j], invF[i - 1]), (1ll * (A / i) * (B / i) % (mod - 1))));
}
tmp = qpow(tmp, C);
res = mul_mod(res, qpow(tmp, mod - 2));
tmp = 1;
for (register int i = 1, j; i <= min(A, C); i = j + 1) {
j = min(A / (A / i), C / (C / i));
tmp = mul_mod(tmp, qpow(mul_mod(F[j], invF[i - 1]), (1ll * (A / i) * (C / i) % (mod - 1))));
}
tmp = qpow(tmp, B);
res = mul_mod(res, qpow(tmp, mod - 2));
return res;
}
inline int calc2(register int A, register int B, register int C) {
register int res = qpow(pmul[A], 1ll * S[B] * S[C] % (mod - 1));
res = mul_mod(res, qpow(pmul[B], 1ll * S[A] * S[C] % (mod - 1)));
register int tmp = 1;
for (register int i = 1, j; i <= min(A, B); i = j + 1) {
j = min(A / (A / i), B / (B / i));
tmp = mul_mod(tmp, qpow(mul_mod(G[j], invG[i - 1]), (1ll * S[A / i] * S[B / i] % (mod - 1))));
}
tmp = qpow(tmp, S[C]);
res = mul_mod(res, qpow(tmp, mod - 2));
tmp = 1;
for (register int i = 1, j; i <= min(A, C); i = j + 1) {
j = min(A / (A / i), C / (C / i));
tmp = mul_mod(tmp, qpow(mul_mod(G[j], invG[i - 1]), (1ll * S[A / i] * S[C / i] % (mod - 1))));
}
tmp = qpow(tmp, S[B]);
res = mul_mod(res, qpow(tmp, mod - 2));
return res;
}
inline int SUM(register int A, register int B) {
register int res = 1;
for (register int i = 1, j; i <= min(A, B); i = j + 1) {
j = min(A / (A / i), B / (B / i));
res = mul_mod(res, qpow(mul_mod(F[j],invF[i - 1]), 1ll * (A / i) * (B / i) % (mod - 1)));
}
return res;
}
inline int calc3(register int A, register int B, register int C) {
register int res = 1;
for (register int i = 1, j; i <= min(min(A, B), C); i = j + 1) {
j = min(min(A / (A / i), B / (B / i)), C / (C / i));
res = mul_mod(res, qpow(mul[B / i], 1ll * (0ll + sumphi[j] - sumphi[i - 1] + (mod - 1)) * (A / i) % (mod - 1) * (C / i) % (mod - 1)));
}
int tmp = 1;
for (register int i = 1, j; i <= min(B, min(A, C)); i = j + 1) {
j = min(min(B / (B / i), A / (A / i)), C / (C / i));
tmp = mul_mod(tmp, qpow(SUM(B / i, A / i), 1ll * (0ll + sumphi[j] - sumphi[i - 1] + (mod - 1)) * (C / i) % (mod - 1)));
}
res = mul_mod(res, qpow(tmp, mod - 2));
for (register int i = 1, j; i <= min(min(A, B), C); i = j + 1) {
j = min(min(A / (A / i), B / (B / i)), C / (C / i));
res = mul_mod(res, qpow(mul[A / i], 1ll * (0ll + sumphi[j] - sumphi[i - 1] + (mod - 1)) * (C / i) % (mod - 1) * (B / i) % (mod - 1)));
}
tmp = 1;
for (register int i = 1, j; i <= min(B, min(A, C)); i = j + 1) {
j = min(min(B / (B / i), A / (A / i)), C / (C / i));
tmp = mul_mod(tmp, qpow(SUM(A / i, C / i), 1ll * (0ll + sumphi[j] - sumphi[i - 1] + (mod - 1)) * (B / i) % (mod - 1)));
}
res = mul_mod(res, qpow(tmp, mod - 2));
return res;
}
int main () {
register int T = read(); mod = read();
init();
register int A, B, C;
while (T--) {
A = read(), B = read(), C = read();
printf("%d %d %d\n", calc1(A, B, C), calc2(A, B, C), calc3(A, B, C));
}
return 0;
}
顺便贴一下ln大法的代码实现：
点击查看代码
inline int calc2(register int A, register int B, register int C) {
int res = 1;
for (int i = 1, j; i <= min(min(A, B), C); i = j + 1) {
j = min(min(A / (A / i), B / (B / i)), C / (C / i));
res = mul_mod(res, qpow(calc0(A / i, B / i, C / i), (0ll + sumphi[j] - sumphi[i - 1] + (mod - 1)) % (mod - 1)));
}
return res;
}
后记
这个题的数学公式也太复杂了...
不得不把公式字体放大，然后又显得太蠢了...
然后卡常吧...
莫比乌斯反演《基础》练习题
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