Bzoj4816: [Sdoi2017]数字表格

题面

戳我

Sol

摆公式：

ans=Πni=1Πmj=1f[gcd(i,j)]

考虑每个gcd的贡献，设n < m

则就是Πnd=1Π⌊nd⌋i=1Π⌊md⌋j=1f[d]∗[gcd(i,j)==1]

=Πnd=1f[d]∑⌊nd⌋i=1∑⌊md⌋j=1[gcd(i,j)==1]

那个指数幂就是∑⌊nd⌋i=1μ[i]∗⌊ndi⌋∗⌊mdi⌋

但指数取模1e9+7显然不行，所以要用到一个欧拉的定理

降幂法

如果直接这样搞加了数论分块也会TLE飞

所以考虑把式子变形一下：di换成k

Πnk=1Πd|kf[d]μ[kd]⌊nk⌋⌊mk⌋

把Πd|kf[d]μ[kd]预处理出来就好 以为可以筛想了好久

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e6 + 1), Zsy(1e9 + 7);

IL ll Read(){
char c = '%'; ll x = 0, z = 1;
for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;
for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
return x * z;
}

int prime[_], num, mu[_], g[_], f[_], s[_];
bool isprime[_];

IL int Pow(RG ll x, RG ll y){
RG ll ret = 1;
for(; y; y >>= 1, x = x * x % Zsy) if(y & 1) ret = ret * x % Zsy;
return ret;
}

IL void Prepare(){
isprime[1] = 1; mu[1] = f[1] = g[1] = s[1] = s[0] = 1;
for(RG int i = 2; i < _; ++i){
if(!isprime[i]){  prime[++num] = i; mu[i] = -1;  }
for(RG int j = 1; j <= num && i * prime[j] < _; ++j){
isprime[i * prime[j]] = 1;
if(i % prime[j]) mu[i * prime[j]] = -mu[i];
else{  mu[i * prime[j]] = 0; break;  }
}
f[i] = (f[i - 1] + f[i - 2]) % Zsy;
g[i] = Pow(f[i], Zsy - 2); s[i] = 1;
}
for(RG int i = 1; i < _; ++i){
if(!mu[i]) continue;
for(RG int j = i, t = 1; j < _; j += i, ++t)
s[j] = 1LL * s[j] * ((mu[i] == 1) ? f[t] : g[t]) % Zsy;
}
for(RG int i = 1; i < _; ++i) s[i] = 1LL * s[i] * s[i - 1] % Zsy;
}

int main(RG int argc, RG char *argv[]){
Prepare();
for(RG ll T = Read(), n, m, ans; T; --T){
n = Read(); m = Read(); ans = 1;
if(n > m) swap(n, m);
for(RG ll k = 1, j; k <= n; k = j + 1){
j = min(n / (n / k), m / (m / k));
ans = 1LL * ans * Pow(1LL * s[j] * Pow(s[k - 1], Zsy - 2) % Zsy, 1LL * (n / k) * (m / k) % (Zsy - 1)) % Zsy;
}
printf("%lld\n", ans);
}
return 0;
}