Topic links: http://acm.hdu.edu.cn/showproblem.php?pid=6588
Title effect: seeking \ (\ sum_ {i = 1} ^ {n} gcd (\ left \ lfloor \ sqrt [3] {i} \ right \ rfloor, i), \ n \ leq 10 ^ {21} \)
Solution: consider \ (\ left \ lfloor \ sqrt [3] {i} \ right \ rfloor \) block, converting the equation into \ (\ sum_ {i = 1} ^ {\ left \ lfloor \ sqrt [ 3] {n} \ right \ rfloor} \ sum_ {j = i ^ 3} ^ {(i + 1) ^ 3-1} gcd (i, j) \)
Consider the function \ (f (n, m) = \ sum_ {i = 1} ^ {m} gcd (n, i) \), since \ (gcd (x, y) = gcd (x \ mod \ y, y ) \), can be derived $$ f (n, m) = \ sum_ {i = 1} ^ {m} gcd (n, i) = \ left \ lfloor \ frac {m} {n} \ right \ rfloor \ cdot f (n, n) + f (n, m \ mod \ n) $$
For values \ (f (n, n) \) may be enumerated by the value of \ (GCD \) derived sum \ (f (n, n) = \ sum_ {d | n} ^ {} d \ cdot \ phi (\ frac {n} {d}) \), the value of the Euler function can be a linear pre-screening \ (O (nlogn) \) all pre \ (f (n, n) \) values
对于\(f(n,m)\)，同样可以得出，\(f(n,m)=\sum_{d|n}^{ } d \sum_{i| \frac{n}{d},i\leq \left \lfloor \frac{m}{d} \right \rfloor}^{ }\mu(i)\cdot\left \lfloor \frac{m}{id} \right \rfloor=\sum_{d|n}^{ } \phi(d)\cdot \left \lfloor \frac{m}{d} \right \rfloor\)
Also, because \ ((i + 1) ^ 3-1 = i ^ 3 + 3i ^ 2 + 3i \) is (I \) multiple \, so for each cycle \ (I \), there $$ \ sum_ {j = i ^ 3} ^ {(i + 1) ^ 3-1} gcd (i, j) = f (i, i ^ 3 + 3i ^ 2 + 3i) -f (i, i ^ 3 ) + gcd (i, i ^ 3) = (3i + 3) \ cdot f (i, i) + i $$
Thus for each interrogation, at most required only once \ (f (n, m) \) like (i.e., the last \ (I \)), for the sake of a \ (f (n, m) \) time complexity degree \ (O (\ sqrt {n}) \), so that each query complexity is \ (O (n + \ sqrt {n}) \). And because a plurality of sets interrogation, offline processing may be considered by the reduced overall time complexity \ (O (nlogn + T \ sqrt {n}) \), where \ (n-\) refers to the \ (10 ^ 7 \)
The topic of people this question was originally card nonlinear approach, but by a certain constant optimization allows \ (O (nlogn) \) practices through
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define N 10000001 4 #define MOD 998244353 5 int n,g[N],phi[N],p[N],cnt,res[15],flg[N]; 6 struct rua 7 { 8 int id; 9 __int128 n; 10 bool operator <(const rua &t)const{return n<t.n;} 11 }a[15]; 12 void read(__int128 &x) 13 { 14 static char ch;static bool neg; 15 for(ch=neg=0;ch<'0' || '9'<ch;neg|=ch=='-',ch=getchar()); 16 for(x=0;'0'<=ch && ch<='9';(x*=10)+=ch-'0',ch=getchar()); 17 x=neg?-x:x; 18 } 19 void pretype() 20 { 21 phi[1]=1; 22 for(int i=2;i<N;i++) 23 { 24 if(!flg[i])p[++cnt]=i,phi[i]=i-1; 25 for(int j=1;j<=cnt && i*p[j]<N;j++) 26 { 27 flg[i*p[j]]=1; 28 if(i%p[j]==0) 29 { 30 phi[i*p[j]]=p[j]*phi[i]; 31 break; 32 } 33 phi[i*p[j]]=(p[j]-1)*phi[i]; 34 } 35 } 36 for(int d=1;d*d<N;d++) 37 for(int n=d*d;n<N;n+=d) 38 { 39 g[n]=(g[n]+1ll*d*phi[n/d])%MOD; 40 if(d*d<n)g[n]=(g[n]+1ll*(n/d)*phi[d])%MOD; 41 } 42 } 43 inline int F(int n,__int128 M) 44 { 45 int res=(M/n)*g[n]%MOD; 46 int m=M%n; 47 if(m==0)return res; 48 for(int i=1;i*i<=n;i++) 49 { 50 int j=n/(n/i); 51 if(n%i==0) 52 { 53 res=(res+1ll*phi[i]*(m/i))%MOD; 54 if(i*i<n)res=(res+1ll*phi[n/i]*(m/(n/i)))%MOD; 55 } 56 i=j; 57 } 58 return res; 59 } 60 int main() 61 { 62 pretype(); 63 scanf("%d",&n); 64 for(int i=1;i<=n;i++) 65 read(a[i].n),a[i].id=i; 66 sort(a+1,a+n+1); 67 int ans=0,j=1; 68 __int128 i=1; 69 while(i*i*i<=a[n].n) 70 { 71 __int128 r=((i+3)*i+3)*i; 72 ans=(ans+MOD-(i*i*g[i]%MOD)+i)%MOD; 73 while(j<=n && a[j].n<r) 74 { 75 res[a[j].id]=(ans+F(i,a[j].n%i)+(a[j].n/i)*g[i])%MOD; 76 j++; 77 } 78 ans=(ans+((i+3)*i+3)*g[i])%MOD; 79 while(j<=n && a[j].n==r) 80 res[a[j++].id]=ans; 81 i++; 82 } 83 for(int i=1;i<=n;i++) 84 printf("%d\n",res[i]); 85 }