一句话题意
求(egin{aligned}prod_{i=1}^nprod_{j=1}^mf_{gcd(i,j)}end{aligned})，多组询问
解题思路
考虑更换枚举(gcd)的方式，由于是连乘(prod)处理起来可能有点抽象，不像(sum)一样非常好理解
(egin{aligned}prod_{i=1}^nprod_{j=1}^mf_{gcd(i,j)}end{aligned})
(egin{aligned}=prod_{p=1}^nf_p^{sum_{i=1}^nsum_{j=1}^m[gcd(i,j)=p]}end{aligned})，这里就是在更换方式枚举(gcd)，对于每一个(n)，由于是(prod)，对于之前的(n)也要再次进行计算
(egin{aligned}=prod_{p=1}^nf_p^{sum_{i=1}^frac npsum_{j=1}^frac mp[gcd(i,j)=1]}end{aligned})，反演老套路，对幂上的柿子进行反演
(egin{aligned}=prod_{p=1}^nf_p^{sum_{i=1}^{frac np}sum_{j=1}^{frac mp}sum_{d|i,d|j}mu(d)}end{aligned})
(egin{aligned}=prod_{p=1}^nf_p^{sum_{d=1}^nmu(d)lfloorfrac n{pd} floorlfloorfrac m{pd} floor}end{aligned})
然后推到到这里按照反演老套路该(dp)化(Q)然后整除分块直接直播切题了，但是怎么化(Q)？？？？？？？歪？
(egin{aligned}prod_{Q=1}^nprod_{d|Q}end{aligned})按照老套路，一定会有这个东西，在思考如何去处理(f)以及(mu)我们更换枚举之后对于枚举(d)的时候不会对(Q)产生影响，把分块的指数提出来，然后改一下下标就可以了
(egin{aligned}=prod_{Q=1}^nleft(prod_{d|Q}f_{d/Q}^{mu(d)} ight)^{lfloorfrac n{Q} floorlfloorfrac m{Q} floor}end{aligned})
对于里面那一堆东西(egin{aligned}prod_{d|Q}f_{d/Q}^{mu(d)}end{aligned})我们直接预处理然后分块计算就行了，复杂度为(O(n+Tlog nsqrt n))
#include<bits/stdc++.h>
#define LL long long
#define _ 0
#define int long long
using namespace std;
/*Grievous Lady*/
template <typename T> void read(T & t){
t = 0;int f = 1;char ch = getchar();
while(ch < '0' || ch > '9'){if(ch == '-')f =- 1;ch = getchar();}
do{t = t * 10 + ch - '0';ch = getchar();}while(ch >= '0' && ch <= '9');t *= f;
}
#define mod 1000000007
const int kato = 1e7 + 10;
const int luna = 1e6;
inline LL quick_pow(LL a , LL b){
LL res = 1;
for(; b ; b >>= 1 , a = a * a % mod){
if(b & 1){
res = res * a % mod;
}
}
return res % mod;
}
int prime[kato] , cnt , f[kato] , invf[kato] , yuni[kato] , mu[kato];
bool ispri[kato];
inline void get_yuni(){
for(int i = 2;i <= luna;i ++){
if(!ispri[i]){
prime[++ cnt] = i;
mu[i] = -1;
}
for(int j = 1;j <= cnt && i * prime[j] <= luna;j ++){
ispri[i * prime[j]] = 1;
if(i % prime[j] == 0){
break;
}
mu[i * prime[j]] = -mu[i];
}
}
for(int i = 2;i <= luna;i ++){
f[i] = (f[i - 1] + f[i - 2]) % mod;
invf[i] = quick_pow(f[i] , mod - 2);
yuni[i] = 1;
}
for(int d = 1;d <= luna;d ++){
if(mu[d] == 1){
for(int Q = d , cnt = 1;Q <= luna;Q += d , cnt ++){
yuni[Q] = yuni[Q] * f[cnt] % mod;
}
}
if(mu[d] == -1){
for(int Q = d , cnt = 1;Q <= luna;Q += d , cnt ++){
yuni[Q] = yuni[Q] * invf[cnt] % mod;
}
}
}
for(int i = 1;i <= luna;i ++){
yuni[i] = yuni[i] * yuni[i - 1] % mod;
}
}
LL n , m , t;
inline int Ame_(){
read(t);
mu[1] = f[1] = invf[1] = yuni[1] = yuni[0] = 1;
get_yuni();
// for(int i = 1;i <= 10;i ++){
// cerr << mu[i] << ' ';
// }
for(; t --> 0 ;){
LL ans = 1;
read(n) , read(m);
if(n > m) swap(n , m);
for(LL l = 1 , r;l <= n;l = r + 1){
r = min(n / (n / l) , m / (m / l));
ans = ans * quick_pow(yuni[r] * quick_pow(yuni[l - 1] , mod - 2) % mod , (n / l) * (m / l)) % mod;
}
printf("%lld
" , ans);
}
// fclose(stdin);
// fclose(stdout);
return ~~(0^_^0);
}
int Ame__ = Ame_();
signed main(){;}