没啥好说的，杜教筛板子题。
[sum_{i=1}^{N} sum_{j=1}^{N}sum_{p=1}^{lfloor frac{N}{j}
floor}sum_{q=1}^{lfloor frac{N}{j}
floor} [gcd(i,j)==1][gcd(p,q)==1]
]
容易发现，我们枚举 (j) 其实是相当于枚举 (gcd)
才不是枚举题目
然后式子可以变成
[sum_{i=1}^{N}sum_{p=1}^{N}sum_{q=1}^{N} [gcd(i,p,q)==1]
]
然后套路式的枚举 (gcd) 依旧不是枚举题目
[sum_{d=1}^{N}sum_{i=1}^{N}sum_{p=1}^{N}sum_{q=1}^{N}[gcd(i,p,q)==d]
]
熟悉的形式，其实就等于
[sum_{d=1}^{N}sum_{i=1}^{lfloor frac{N}{d}
floor} sum_{p=1}^{lfloor frac{N}{d}
floor} sum_{q=1}^{lfloor frac{N}{d}
floor} mu(d)
]
[sum_{d=1}^{N} mu(d) lfloor frac{N}{d}
floor^3
]
然后整除分块就完了，由于 (N) 比较大，大力杜教筛就完事了，话说我好像是这题除掉出题人的最优解
// powered by c++11
// by Isaunoya
#include<bits/stdc++.h>
#define rep(i , x , y) for(register int i = (x) ; i <= (y) ; ++ i)
#define Rep(i , x , y) for(register int i = (x) ; i >= (y) ; -- i)
using namespace std ;
using db = double ;
using ll = long long ;
using uint = unsigned int ;
#define int long long
using pii = pair < int , int > ;
#define ve vector
#define Tp template
#define all(v) v.begin() , v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
#define fir first
#define sec second
// the cmin && cmax
Tp < class T > void cmax(T & x , const T & y) { if(x < y) x = y ; }
Tp < class T > void cmin(T & x , const T & y) { if(x > y) x = y ; }
// sort , unique , reverse
Tp < class T > void sort(ve < T > & v) { sort(all(v)) ; }
Tp < class T > void unique(ve < T > & v) { sort(all(v)) ; v.erase(unique(all(v)) , v.end()) ; }
Tp < class T > void reverse(ve < T > & v) { reverse(all(v)) ; }
const int SZ = 0x191981 ;
struct FILEIN {
~ FILEIN () {} char qwq[SZ] , * S = qwq , * T = qwq , ch ;
char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq , 1 , SZ , stdin) , S == T) ? EOF : * S ++ ; }
FILEIN & operator >> (char & c) { while(isspace(c = GETC())) ; return * this ; }
FILEIN & operator >> (string & s) {
while(isspace(ch = GETC())) ; s = ch ;
while(! isspace(ch = GETC())) s += ch ; return * this ;
}
Tp < class T > void read(T & x) {
bool sign = 1 ; while((ch = GETC()) < 0x30) if(ch == 0x2d) sign = 0 ;
x = (ch ^ 0x30) ; while((ch = GETC()) > 0x2f) x = x * 0xa + (ch ^ 0x30) ;
x = sign ? x : -x ;
}
FILEIN & operator >> (int & x) { return read(x) , * this ; }
FILEIN & operator >> (signed & x) { return read(x) , * this ; }
FILEIN & operator >> (unsigned & x) { return read(x) , * this ; }
} in ;
struct FILEOUT { const static int LIMIT = 0x114514 ;
char quq[SZ] , ST[0x114] ; signed sz , O ;
~ FILEOUT () { sz = O = 0 ; }
void flush() { fwrite(quq , 1 , O , stdout) ; fflush(stdout) ; O = 0 ; }
FILEOUT & operator << (char c) { return quq[O ++] = c , * this ; }
FILEOUT & operator << (string str) {
if(O > LIMIT) flush() ; for(char c : str) quq[O ++] = c ; return * this ;
}
Tp < class T > void write(T x) {
if(O > LIMIT) flush() ; if(x < 0) { quq[O ++] = 0x2d ; x = -x ; }
do { ST[++ sz] = x % 0xa ^ 0x30 ; x /= 0xa ; } while(x) ;
while(sz) quq[O ++] = ST[sz --] ; return ;
}
FILEOUT & operator << (int x) { return write(x) , * this ; }
FILEOUT & operator << (signed x) { return write(x) , * this ; }
FILEOUT & operator << (unsigned x) { return write(x) , * this ; }
} out ;
const int maxn = 5e5 ;
int mu[maxn + 10] ;
const int mod = 998244353 ;
map < int , int > _mu ;
int getmu(int x) {
if(x <= maxn) return mu[x] ;
if(_mu[x]) return _mu[x] ;
int ans = 1 ;
int l = 2 , r = 0 ;
for( ; l <= x ; l = r + 1) {
r = x / (x / l) ;
ans -= getmu(x / l) * (r - l + 1) ;
ans = (ans + mod) % mod ;
}
return _mu[x] = ans ;
}
signed main() {
#ifdef _WIN64
freopen("testdata.in" , "r" , stdin) ;
#else
ios_base :: sync_with_stdio(false) ;
cin.tie(nullptr) , cout.tie(nullptr) ;
#endif
// code begin.
mu[1] = 1 ;
for(int i = 1 ; i <= maxn ; i ++)
for(int j = i + i ; j <= maxn ; j += i)
mu[j] -= mu[i] ;
for(int i = 2 ; i <= maxn ; i ++)
mu[i] = (mu[i] + mu[i - 1]) % mod ;
int n ;
in >> n ;
int l = 1 , r = 0 ;
int ans = 0 ;
for( ; l <= n ; l = r + 1) {
r = n / (n / l) ;
int qwq = (n / l) * (n / l) % mod * (n / l) % mod ;
ans = (ans + (getmu(r) - getmu(l - 1) + mod) % mod * qwq % mod) % mod ;
}
out << ans << '
' ;
return out.flush() , 0 ;
// code end.
}