BZOJ 4816.「SDOI2017」数字表格

Very obviously, 题目让我们求 i=1nj=1mfgcd(i,j)\prod\limits_{i = 1}^n \prod\limits_{j = 1}^m f_{\gcd(i,j)}
按照套路,我们枚举 gcd\gcd,转化为求 i=1min(n,m)fix=1ny=1m[gcd(x,y)=i]\prod\limits_{i = 1}^{\min(n,m)} {f_i}^{\sum\limits_{x = 1}^n \sum\limits_{y = 1}^m [\gcd(x,y) = i]}

把指数那个式子拉出来单独讨论。
根据套路,有

那么

显然外层的指数可以数论分块,内层这个东西好像不太可做……
那就暴力预处理前缀积啊,反正枚举倍数 O(nlogn)O(n \log n)

另外此题超卡常,请全开 int 并预处理前缀积的逆元。

代码:

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#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1e6;
const int mod = 1e9 + 7;
int t;
int cnt,prime[N + 5],vis[N + 5],mu[N + 5];
int f[N + 5],g[N + 5],prod[N + 5],inv[N + 5],ans;
int fpow(int a,int b)
{
int ret = 1;
for(;b;b >>= 1)
{
if(b & 1)
ret = (long long)ret * a % mod;
a = (long long)a * a % mod;
}
return ret;
}
int main()
{
f[1] = g[1] = mu[1] = prod[0] = inv[0] = prod[1] = inv[1] = 1;
for(register int i = 2;i <= N;++i)
{
prod[i] = 1,f[i] = (f[i - 1] + f[i - 2]) % mod,g[i] = fpow(f[i],mod - 2);
if(!vis[i])
prime[++cnt] = i,mu[i] = -1;
for(register int j = 1;j <= cnt && i * prime[j] <= N;++j)
{
vis[i * prime[j]] = 1;
if(!(i % prime[j]))
break;
mu[i * prime[j]] = -mu[i];
}
}
for(register int i = 1;i <= N;++i)
if(mu[i])
for(register int j = 1;i * j <= N;++j)
prod[i * j] = (long long)prod[i * j] * (mu[i] == 1 ? f[j] : g[j]) % mod;
for(register int i = 1;i <= N;++i)
prod[i] = (long long)prod[i] * prod[i - 1] % mod,inv[i] = fpow(prod[i],mod - 2);
scanf("%d",&t);
int n,m;
while(t--)
{
ans = 1;
scanf("%d%d",&n,&m);
if(n > m)
swap(n,m);
for(register int l = 1,r;l <= n;l = r + 1)
{
r = min(n / (n / l),m / (m / l));
ans = (long long)ans * fpow((long long)prod[r] * inv[l - 1] % mod,(long long)(n / l) * (m / l) % (mod - 1)) % mod;
}
printf("%d\n",ans);
}
}