https://nanti.jisuanke.com/t/41300
题意:求\(\sum_{i=1}^n\phi(i)\phi(j)2^{\phi(i)\phi(j)}\)
\(f_i=\sum_{k=1}^n[\phi(k)==i]\)
\(\sum_{i=1}^n\phi(i)\phi(j)2^{\phi(i)\phi(j)}\)
\(=\sum_{i=1}^n\sum_{j=1}^nf_if_jij2^{ij}\)
\(=2\sum_{i=1}^n\sum_{j=1}^if_if_jij2^{ij}-\sum_{i=1}^nf_ii2^{i^2}\)
\(=2\sum_{i=1}^nif_i\sum_{j=1}^ijf_j2^{i^2+j^2-(i-j)^2}-\sum_{i=1}^nf_ii2^{i^2}\)
\(=2\sum_{i=1}^nif_i{\sqrt 2}^{i^2}\sum_{j=1}^ijf_j{\sqrt 2}^{j^2}{\sqrt 2}^{-(i-j)^2}-\sum_{i=1}^nf_ii2^{i^2}\)
后一个\(\sum\)ntt,预处理f,\(\sqrt2\)的二次剩余
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include
//#include
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define mt make_tuple
#define pi acos(-1.0)
#define ll long long
#define vi vector
#define mod 998244353
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair
#define pil pair
#define pli pair
#define pii pair
#define ull unsigned long long
#define bpc __builtin_popcount
#define base 1000000000000000000ll
#define fin freopen("1.in","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
#define mr mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
templateinline T const& MAX(T const &a,T const &b){return a>b?a:b;}
templateinline T const& MIN(T const &a,T const &b){return a>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=mul(ans,a,c);a=mul(a,a,c),b>>=1;}return ans;}
using namespace std;
//using namespace __gnu_pbds;
const ull ba=233;
const db eps=1e-5;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=2000000+10,inf=0x3f3f3f3f;
int prime[N],cnt,phi[N],f[N];
bool mark[N];
void init()
{
phi[1]=1;
for(int i=2;i>1]>>1) | ((i&1)<<(bit-1));
}
void ntt(ll *a,int n,int dft)
{
for(int i=0;i