题目链接
Description
![](https://img.laitimes.com/img/_0nNw4CM6IyYiwiM6ICdiwiIwczLcVmds92czlGZvwVP9EUTDZ0aRJkSwk0LcxGbpZ2LcBDM08CXlpXazRnbvZ2LcRlMMVDT2EWNvwFdu9mZvwlb1IjYwQWbiZnUzgVMkNDTwYVbiVHNHpleO1GTulzRilWO5x0LcRHelR3LcJzLctmch1mclRXY39TMzgTN1gDN0ETMyMDM4EDMy8CX0Vmbu4GZzNmLn9Gbi1yZtl2Lc9CX6MHc0RHaiojIsJye.jpg)
分析:
∏i=1n∏j=1mf(gcd(i,j)) ∏ i = 1 n ∏ j = 1 m f ( g c d ( i , j ) )
对于这种式子,一般我们会枚举gcd(每个f值要使用几次)
设 g(d)=∑ni=1∑mj=1[gcd(i,j)=d] g ( d ) = ∑ i = 1 n ∑ j = 1 m [ g c d ( i , j ) = d ]
∏d=1min(n,m)f(d)g(d) ∏ d = 1 m i n ( n , m ) f ( d ) g ( d )
把 ∑ni=1∑mj=1[gcd(i,j)=d] ∑ i = 1 n ∑ j = 1 m [ g c d ( i , j ) = d ] 用反演化开
g(d)=∑k=1min(n/d,m/d)nkdmkdμ(k) g ( d ) = ∑ k = 1 m i n ( n / d , m / d ) n k d m k d μ ( k )
∏d=1min(n,m)f(d)∑k=1nkdmkdμ(k) ∏ d = 1 m i n ( n , m ) f ( d ) ∑ k = 1 n k d m k d μ ( k )
像我这种naive的选手,到这里就直接做了
继续,我们看看 g(d) g ( d ) 能不能进一步优化,设 T=kd T = k d
g(d)=∑k=1min(n/d,m/d)nkdmkdμ(k)=∑d|TnTmTμ(Td) g ( d ) = ∑ k = 1 m i n ( n / d , m / d ) n k d m k d μ ( k ) = ∑ d | T n T m T μ ( T d )
这样我们就可把 T T 的枚举拉出来
∏d=1min(n,m)f(d)∑d|TnTmTμ(Td)∏d=1min(n,m)f(d)∑d|TnTmTμ(Td)
∏d=1min(n,m)∏d|Tf(d)nTmTμ(Td) ∏ d = 1 m i n ( n , m ) ∏ d | T f ( d ) n T m T μ ( T d )
枚举顺序换一下
∏T=1min(n,m)∏d|Tf(d)nTmTμ(Td)=∏T=1min(n,m)(∏d|Tf(d)μ(Td))nTmT ∏ T = 1 m i n ( n , m ) ∏ d | T f ( d ) n T m T μ ( T d ) = ∏ T = 1 m i n ( n , m ) ( ∏ d | T f ( d ) μ ( T d ) ) n T m T
发现对于每个 T T ,∏d|Tf(d)μ(Td)∏d|Tf(d)μ(Td)的值是固定的,与 n n 和mm无关,于是我们先用筛法预处理出每个 T T 对应的这个式子的值,前缀积一下
而外面这一层∏min(n,m)T=1()nTmT∏T=1min(n,m)()nTmT可以分块处理
所以时间复杂度: O((max(n,m)+T(n−−√+m−−√))logMOD) O ( ( m a x ( n , m ) + T ( n + m ) ) l o g M O D )
tip
∏d|Tf(d)μ(Td) ∏ d | T f ( d ) μ ( T d ) 直接暴力处理即可
设 sum s u m 为前缀积,注意 sum[0]=1 s u m [ 0 ] = 1
一开始预处理就慢到爆,怀疑人生
KSM次数多了也费时,所以我们预处理f的逆元,减少KSM的调用
因为要计算 f(d)μ(Td) f ( d ) μ ( T d ) ,而 μ μ <script type="math/tex" id="MathJax-Element-1569">μ</script>的取值只有-1,1,所以预处理f的逆元反而比较方便
多%防爆ll
还是免不了T掉了
ll改int,计算过程中强转ll
时间就少了一半诶
#include<cstdio>
#include<cstring>
#include<iostream>
#define ll long long
using namespace std;
const int N=;
const ll p=e9+;
int sshu[N+],tot=,mu[N+],n,m;
int f[N+],sum[N+],inv[N+];
bool no[N+];
inline int KSM(int a,int b) {
int t=;
a%=p;
while (b) {
if (b&) t=(LL*t*a)%p;
b>>=;
a=(LL*a*a)%p;
}
return t%p;
}
void prepare() {
mu[]=;
for (int i=;i<=N;i++) {
if (!no[i]) {
sshu[++tot]=i;
mu[i]=-;
}
for (int j=;j<=tot&&sshu[j]*i<=N;j++) {
no[sshu[j]*i]=;
if (i%sshu[j]==) {
mu[sshu[j]*i]=;
break;
}
mu[sshu[j]*i]=-mu[i];
}
}
f[]=; f[]=;
sum[]=; sum[]=; //前缀积
inv[]=; //f的逆元
for (int i=;i<=N;i++) f[i]=(f[i-]+f[i-])%p,inv[i]=KSM(f[i],p-),sum[i]=;
for (int i=;i<=N;i++)
if (mu[i]!=) {
for (int j=;j*i<=N;j++) // j*i=T i=T/d j=d
if (mu[i]>) sum[j*i]=(LL*sum[j*i]*f[j])%p;
else sum[j*i]=(LL*sum[j*i]*inv[j])%p;
}
for (int i=;i<=N;i++) sum[i]=(LL*sum[i]*sum[i-])%p;
}
int main()
{
prepare();
int T;
scanf("%d",&T);
while (T--) {
scanf("%d%d",&n,&m);
int last,ans=;
for (int i=;i<=min(n,m);i=last+) {
last=min(n/(n/i),m/(m/i));
ans=LL*ans%p*KSM(LL*sum[last]*KSM(sum[i-],p-)%p,LL*(n/i)*(m/i)%(p-))%p;
}
printf("%d\n",ans%p);
}
return ;
}