[SDOI2017] 数字表格
Posted qrsikno
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还是一样, 我们应用枚举GCD的套路:
[
Ans = prod_{i}prod_{j} f(gcd(i, j)) = prod_{d} f(d)^ {sum_{i}sum_{j} [(i, j) = d]}= prod_{d} f(d)^ {sum_{s} mu(s)frac{n}{ds}frac{m}{ds}} \]
之后我们继续应用合并元的套路:
[
Ans = prod_{d} prod_{i | d} f(i) ^ {mu(frac{d}{i})frac{n}{d}frac{m}{d}} \]
之后考虑整除分块, 因为要对(a^{frac{n}{d}frac{m}{d}})整除分块,所以必须把所有其他项视为整体(g(n))
[
= prod_{d} f(i) ^ {sum_{i | d}mu(frac{d}{i})frac{n}{d}frac{m}{d}} = prod_{d} (prod_{i | d} f(i) ^ {mu(frac{d}{i})}) ^ {frac{n}{d}frac{m}{d}} Let ~ g(n) = prod_{d | n} f(d) ^ {mu(frac{n}{d})} Ans = prod_{d = 1}^{min(n, m)} g(d) ^ {frac{n}{d} frac{m}{d}}\]
发现(n)的范围很小,于是我们可以枚举约数/倍数暴力计算. 这里枚举倍数的复杂度较小,所以我们采用枚举倍数.
然后就做完了,总复杂度:[O(Tsqrt n ~log_{2}Mod + n ~ln~n + n log_2n)]
Code
#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
#define drep(i, a, b) for(int i = (a), i##_end_ = (b); i >= i##_end_; --i)
#define clar(a, b) memset((a), (b), sizeof(a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
typedef long long LL;
typedef long double LD;
int read() {
char ch = getchar();
int x = 0, flag = 1;
for (;!isdigit(ch); ch = getchar()) if (ch == ‘-‘) flag *= -1;
for (;isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
return x * flag;
}
void write(int x) {
if (x < 0) putchar(‘-‘), x = -x;
if (x >= 10) write(x / 10);
putchar(x % 10 + 48);
}
const int Maxn = 1e6 + 9, Mod = 1000000007;
LL fpm(LL base, LL tims) {
LL r = 1;
for (; tims; tims >>= 1) {
if (tims & 1) r = 1ll * r * base % Mod;
base = 1ll * base * base % Mod;
}
return r;
}
static int prime[Maxn], tot, mu[Maxn];
static bool isnprime[Maxn];
void linearSieve() {
mu[1] = 1;
rep (i, 2, Maxn - 1) {
if (!isnprime[i]) prime[++tot] = i, mu[i] = -1;
for (int k, j = 1; j <= tot && (k = prime[j] * i) < Maxn; ++j) {
isnprime[k] = 1;
if (i % prime[j] == 0) {
mu[k] = 0;
break;
} else mu[k] = -mu[i];
}
}
}
static int Fib[Maxn], invFib[Maxn];
static int F[Maxn], prodF[Maxn], invProdF[Maxn];
void init() {
linearSieve();
Fib[0] = 0, Fib[1] = 1;
rep (i, 2, Maxn - 1) Fib[i] = (Fib[i - 1] + Fib[i - 2]) % Mod;
rep (i, 1, Maxn - 1) invFib[i] = fpm(Fib[i], Mod - 2), F[i] = 1;
rep (i, 1, Maxn - 1)
for (int j = i; j < Maxn; j += i) {
if (mu[j / i] == -1) F[j] = 1ll * F[j] * invFib[i] % Mod;
if (mu[j / i] == 1) F[j] = 1ll * F[j] * Fib[i] % Mod;
}
prodF[0] = invProdF[0] = 1;
rep (i, 1, Maxn - 1) {
prodF[i] = prodF[i - 1] * 1ll * F[i] % Mod;
invProdF[i] = fpm(prodF[i], Mod - 2);
}
}
void solve() {
int T = read();
while (T--) {
int n = read(), m = read();
int Limit = min(n, m); LL ans = 1;
for (int l = 1, r; l <= Limit; l = r + 1) {
r = min(n / (n / l), m / (m / l));
ans = ans * 1ll * fpm(1ll * prodF[r] * invProdF[l - 1] % Mod, 1ll * (n / l) * (m / l)) % Mod;
}
printf("%lld
", ans);
}
}
int main() {
freopen("loj2000.in", "r", stdin);
freopen("loj2000.out", "w", stdout);
init();
solve();
#ifdef Qrsikno
debug("
Running time: %.3lf(s)
", clock() * 1.0 / CLOCKS_PER_SEC);
#endif
return 0;
}
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