P7486 「Stoi2031」彩虹
作者：互联网
P7486 「Stoi2031」彩虹
令 S ( a , b ) = ∏ i = 1 a ∏ j = 1 b lcm ( a , b ) lcm ( a , b ) S(a,b)=\prod\limits_{i=1}^{a}\prod\limits_{j=1}^{b}\operatorname{lcm}(a,b)^{\operatorname{lcm}(a,b)} S(a,b)=i=1∏aj=1∏blcm(a,b)lcm(a,b)。
求：
S
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S
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\frac{S(r,r)\times S(l-1,l-1)}{S(r,l-1)\times S(l-1,r)}
S(r,l−1)×S(l−1,r)S(r,r)×S(l−1,l−1)
不妨设
a
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a \leq b
a≤b。
令
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dp=t
dp=t，则有：
S
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lcm
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lcm
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gcd
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gcd
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gcd
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μ
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\begin{aligned} S(a,b)&=\prod\limits_{i=1}^{a}\prod\limits_{j=1}^{b}\operatorname{lcm}(a,b)^{\operatorname{lcm}(a,b)}\\ &=\prod_{i=1}^{a}\prod_{j=1}^{b}\frac{ij}{\gcd(a,b)}^{(\frac{ij}{\gcd(a,b)})}\\ &=\prod_{d=1}^{a}\prod_{i=1}^{a}\prod_{j=1}^{b}\frac{ij}{d}^{(\frac{ij}{d})[\gcd(i,j)=d]}\\ &=\prod_{d=1}^{a}\prod_{i=1}^{\lfloor\frac{a}{d}\rfloor}\prod_{j=1}^{\lfloor\frac{b}{d}\rfloor}(ijd)^{ijd[\gcd(i,j)=1]}\\ &=\prod_{d=1}^{a}\prod_{i=1}^{\lfloor\frac{a}{d}\rfloor}\prod_{j=1}^{\lfloor\frac{b}{d}\rfloor}(ijd)^{ijd\sum\limits_{p=1}^{\lfloor\frac{a}{d}\rfloor}\mu(p)[p|i][p|j]}\\ &=\prod_{d=1}^{a}\prod_{p=1}^{\lfloor\frac{a}{d}\rfloor}\prod_{i=1}^{\lfloor\frac{a}{d}\rfloor}[p|i]\prod_{j=1}^{\lfloor\frac{b}{d}\rfloor}[p|j](ijd)^{ijd\mu(p)}\\ &=\prod_{d=1}^{a}\prod_{p=1}^{\lfloor\frac{a}{d}\rfloor}\prod_{i=1}^{\lfloor\frac{a}{dp}\rfloor}\prod_{j=1}^{\lfloor\frac{b}{dp}\rfloor}(ijdp^2)^{ijdp^2\mu(p)}\\ &=\prod_{t=1}^{a}\prod_{p|t}\prod_{i=1}^{\lfloor\frac{a}{t}\rfloor}\prod_{j=1}^{\lfloor\frac{b}{t}\rfloor}(ijtp)^{ijtp\mu(p)}\\ \end{aligned}
S(a,b)=i=1∏aj=1∏blcm(a,b)lcm(a,b)=i=1∏aj=1∏bgcd(a,b)ij(gcd(a,b)ij)=d=1∏ai=1∏aj=1∏bdij(dij)[gcd(i,j)=d]=d=1∏ai=1∏⌊da⌋j=1∏⌊db⌋(ijd)ijd[gcd(i,j)=1]=d=1∏ai=1∏⌊da⌋j=1∏⌊db⌋(ijd)ijdp=1∑⌊da⌋μ(p)[p∣i][p∣j]=d=1∏ap=1∏⌊da⌋i=1∏⌊da⌋[p∣i]j=1∏⌊db⌋[p∣j](ijd)ijdμ(p)=d=1∏ap=1∏⌊da⌋i=1∏⌊dpa⌋j=1∏⌊dpb⌋(ijdp2)ijdp2μ(p)=t=1∏ap∣t∏i=1∏⌊ta⌋j=1∏⌊tb⌋(ijtp)ijtpμ(p)
令
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s(x)=\sum\limits_{i=1}^{x}i=\frac{x(x+1)}{2}
s(x)=i=1∑xi=2x(x+1)，
f
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f(x)=\prod\limits_{i=1}^{x}i^i
f(x)=i=1∏xii，则有：
∏
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\prod_{i=1}^{n}\prod_{j=1}^{m}(ij)^{ij}=f(n)^{s(m)}\times f(m)^{s(n)}
i=1∏nj=1∏m(ij)ij=f(n)s(m)×f(m)s(n)
令其为
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G(n,m)
G(n,m)。
另有：
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\prod_{i=1}^{n}\prod_{j=1}^{m}(tp)^{ij}=(tp)^{s(n)\times S(m)}
i=1∏nj=1∏m(tp)ij=(tp)s(n)×S(m)
带回原式有：
∏
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\prod_{t=1}^{a}\left( \prod_{p|t}\left( G(\lfloor\frac{a}{t}\rfloor,\lfloor\frac{b}{t}\rfloor) \times (tp)^{s(\lfloor\frac{a}{t}\rfloor)\times s(\lfloor\frac{b}{t}\rfloor)} \right)^{p\mu(p)} \right)^{t}
t=1∏a⎝⎛p∣t∏(G(⌊ta⌋,⌊tb⌋)×(tp)s(⌊ta⌋)×s(⌊tb⌋))pμ(p)⎠⎞t
另
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h(x)=\sum\limits_{d|x}d\mu(d),y(x)=\prod\limits_{d|x}d^{d\mu(d)}
h(x)=d∣x∑dμ(d),y(x)=d∣x∏ddμ(d)，则有：
∏
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⋅
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\prod_{t=1}^{a}G(\lfloor\frac{a}{t}\rfloor,\lfloor\frac{b}{t}\rfloor)^{t \cdot h(t)}\times (t^{h(t)}\times y(t))^{t\times s(\lfloor\frac{a}{t}\rfloor)\times s(\lfloor\frac{b}{t}\rfloor)}
t=1∏aG(⌊ta⌋,⌊tb⌋)t⋅h(t)×(th(t)×y(t))t×s(⌊ta⌋)×s(⌊tb⌋)
再令
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hh(x)=x\times h(x),yy(x)=(x^{h(x)}\times y(x))^{x}
hh(x)=x×h(x),yy(x)=(xh(x)×y(x))x，可得：
∏
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\prod_{t=1}^{a}G(\lfloor\frac{a}{t}\rfloor,\lfloor\frac{b}{t}\rfloor)^{hh(t)}\times yy(t)^{s(\lfloor\frac{a}{t}\rfloor)\times s(\lfloor\frac{b}{t}\rfloor)}
t=1∏aG(⌊ta⌋,⌊tb⌋)hh(t)×yy(t)s(⌊ta⌋)×s(⌊tb⌋)
最后前缀和
+
\text{+}
+ 前缀积
+
\text{+}
+ 数论分块即可。
再加一个小优化，设一个阈值 S S S，对于 1 ≤ t < S 1 \leq t < S 1≤t<S 的直接暴力算，因为这部分 l , r l,r l,r 相差较小。
对于 S ≤ t ≤ n S \leq t \leq n S≤t≤n 的，再数论分块算。
#include <bits/stdc++.h>
using namespace std;
inline int read()
{
int x = 0, f = 1;
char c = getchar();
while(c < '0' || c > '9')
{
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
{
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
inline void write(int x)
{
if(x < 0)
{
putchar('-');
x = -x;
}
if(x > 9)
write(x / 10);
putchar(x % 10 + '0');
}
const int _ = 1e6 + 7, mod = 32465177;
bool vis[_];
int cnt, pr[_], mu[_], f[_], h[_], y[_];
int ksm(int x, int y)
{
int res = 1;
while(y)
{
if(y & 1) res = 1ll * res * x % mod;
x = 1ll * x * x % mod;
y >>= 1;
}
return res;
}
void init()
{
mu[1] = 1;
f[1] = 1;
y[1] = 1;
for(int i = 2; i <= _ - 7; ++i)
{
f[i] = 1ll * f[i - 1] * ksm(i, i) % mod;
y[i] = 1;
if(!vis[i])
{
pr[++cnt] = i;
mu[i] = -1;
}
for(int j = 1; j <= cnt && i * pr[j] <= _ - 7; ++j)
{
int p = i * pr[j];
vis[p] = 1;
if(i % pr[j] == 0)
{
mu[p] = 0;
break;
}
mu[p] = -mu[i];
}
}
for(int i = 1; i <= _ - 7; ++i)
for(int j = 1; i * j <= _ - 7; ++j)
{
h[i * j] = ((h[i * j] + mu[i] * i) % (mod - 1) + (mod - 1)) % (mod - 1);
y[i * j] = 1ll * y[i * j] * ksm(i, i * mu[i] + mod - 1) % mod;
}
for(int i = 1; i <= _ - 7; ++i)
{
y[i] = ksm(1ll * ksm(i, h[i]) * y[i] % mod, i);
h[i] = 1ll * h[i] * i % (mod - 1);
}
y[0] = 1;
for(int i = 2; i <= _ - 7; ++i)
{
y[i] = 1ll * y[i - 1] * y[i] % mod;
h[i] = (h[i] + h[i - 1]) % (mod - 1);
}
}
int s(int x)
{
return 1ll * x * (x + 1) / 2 % (mod - 1);
}
int g(int x, int y)
{
return 1ll * ksm(f[x], s(y)) * ksm(f[y], s(x)) % mod;
}
int hh(int l, int r)
{
return ((h[r] - h[l - 1]) % (mod - 1) + (mod - 1)) % (mod - 1);
}
int yy(int l, int r)
{
return 1ll * y[r] * ksm(y[l - 1], mod - 2) % mod;
}
int S(int a, int b)
{
if(a > b) swap(a, b);
int res = 1;
for(int i = 1, j; i <= a; i = j + 1)
{
j = min(a / (a / i), b / (b / i));
res = 1ll * res * ksm(g(a / i, b / i), hh(i, j)) % mod * ksm(yy(i, j), 1ll * s(a / i) * s(b / i) % (mod - 1)) % mod;
}
return res;
}
int solve(int l, int r)
{
swap(l, r);
int ans1 = 1ll * S(r, r) * S(l - 1, l - 1) % mod, ans2 = 1ll * S(r, l - 1) * S(l - 1, r) % mod;
return 1ll * ans1 * ksm(ans2, mod - 2) % mod;
}
signed main()
{
init();
int t = read(), n = read();
while(t--)
{
write(solve(read(), read()));
putchar('\n');
}
}
标签：彩虹,lfloor,P7486,frac,Stoi2031,int,rfloor,prod,mod 来源： https://blog.csdn.net/qq_46258139/article/details/122552662