P3704 [SDOI2017]數字表格 題解
阿新 • • 發佈:2022-04-17
一道莫反套路題,其思路基本貫穿幽靈樂團。
下面欽定 \(n \le m\),這樣做答案不變,規定所有除法都是整除(所以不要亂約分)。
\[\prod_{i=1}^{n}\prod_{j=1}^{m}f_{\gcd(i,j)} \]列舉 \(\gcd(i,j)\),注意這一步之後 \([gcd(i,j)=d]\) 是在指數上的。
\[\prod_{d=1}^{n}\prod_{i=1}^{n}\prod_{j=1}^{m}f_{d}^{[\gcd(i,j)=d]} \]然後將後面兩個 \(\prod\) 丟到指數上。
\[\prod_{d=1}^{n}f_d^{\sum_{i=1}^{n}\sum_{j=1}^{m}[\gcd(i,j)=d]} \]然後指數上就是一個熟悉的形式(不會的見文末),得到:
套路式的,列舉 \(T=td\) 和 \(d \mid T\),同時將指數上的 \(\sum\) 拉下來,價格括號,就有:
\[\prod_{T=1}^{n}(\prod_{d \mid T}f_d^{\mu(\frac{T}{d})})^{\frac{n}{T} \times \frac{m}{T}} \]然後中間括號括起來的這部分是可以預處理的,複雜度是調和級數 \(O(\sum_{i=1}^{n}\dfrac{n}{i})\)
注意指數開 long long
。
GitHub:CodeBase-of-Plozia
Code:
/* ========= Plozia ========= Author:Plozia Problem:P3704 [SDOI2017]數字表格 Date:2022/3/9 ========= Plozia ========= */ #include <bits/stdc++.h> typedef long long LL; const int MAXN = 1e6 + 5; const LL P = 1e9 + 7; int t, n, m, mu[MAXN], Prime[MAXN], cnt_Prime; LL f[MAXN], sum[MAXN], inv[MAXN]; bool book[MAXN]; int Read() { int sum = 0, fh = 1; char ch = getchar(); for (; ch < '0' || ch > '9'; ch = getchar()) fh -= (ch == '-') << 1; for (; ch >= '0' && ch <= '9'; ch = getchar()) sum = sum * 10 + (ch ^ 48); return sum * fh; } int Max(int fir, int sec) { return (fir > sec) ? fir : sec; } int Min(int fir, int sec) { return (fir < sec) ? fir : sec; } LL ksm(LL a, LL b = P - 2, LL p = P) { if (b == -1) b = P - 2; LL s = 1 % p; for (; b; b >>= 1, a = a * a % p) if (b & 1) s = s * a % p; return s; } void init() { f[1] = 1; book[1] = mu[1] = 1; inv[1] = 1; int tmp = MAXN - 5; for (int i = 2; i <= tmp; ++i) f[i] = (f[i - 1] + f[i - 2]) % P, inv[i] = ksm(f[i]); for (int i = 2; i <= tmp; ++i) { if (!book[i]) { Prime[++cnt_Prime] = i; mu[i] = -1; } for (int j = 1; j <= cnt_Prime; ++j) { if (i * Prime[j] > tmp) break ; book[i * Prime[j]] = 1; if (i % Prime[j] == 0) { mu[i * Prime[j]] = 0; break ; } mu[i * Prime[j]] = -mu[i]; } } for (int i = 1; i <= tmp; ++i) sum[i] = 1; for (int i = 1; i <= tmp; ++i) { if (!mu[i]) continue ; for (int j = i; j <= tmp; j += i) sum[j] = sum[j] * ((mu[i] == 1) ? f[j / i] : inv[j / i]) % P; } sum[0] = 1; inv[0] = 1; for (int i = 1; i <= tmp; ++i) sum[i] = sum[i - 1] * sum[i] % P; for (int i = 1; i <= tmp; ++i) inv[i] = ksm(sum[i]); } int main() { t = Read(); init(); while (t--) { n = Read(), m = Read(); LL ans = 1; if (n > m) std::swap(n, m); for (int l = 1, r; l <= n; l = r + 1) { r = Min(n / (n / l), m / (m / l)); ans = ans * ksm(sum[r] * inv[l - 1] % P, 1ll * (n / l) * (m / l)) % P; } printf("%lld\n", ans); } return 0; }
關於 \(\sum_{i=1}^{n}\sum_{j=1}^{m}[\gcd(i,j)=d]\) 的化簡(第一步除法不是整除,後面的都是):
\[\sum_{i=1}^{n}\sum_{j=1}^{m}[\gcd(i/d,j/d)=1] \] \[\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}[\gcd(i,j)=1] \] \[\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}\sum_{t \mid \gcd(i,j)=1}\mu(t) \] \[\sum_{t=1}^{n/d}\mu(t)\sum_{i=1}^{n/d}\sum_{j=1}^{n/d}[t \mid \gcd(i,j)] \] \[\sum_{t=1}^{n/d}\dfrac{n}{td}\dfrac{m}{td} \]這還看不懂請重修莫反以及推式子。