Bzoj 2693 Jzptab "The inverse of the Mo"

Description

Hint

T <= 10000
N, m<=10000000

Solution

Almost the same as the Bzoj 2154 number table, except that the question becomes multiple groups, and the complexity of the previous one is too much.

Still write the answer

and two more enumerators.

We try to change the sum of the indicator + prefix and continue to subtract an enumerator

And then there's the

For this thing we define it as H (D)

Enables prefixes and preprocessing.

Consider the multiple of enumerations I and I

But this kind of treatment is obviously not acceptable.

It seems that only the complexity of O (n) can be accepted

Can you put h in a linear sieve and handle it?

For a prime p, its new μ value is obviously p-p^2.

If P is the product of the first term of multiple primes

Obviously H is the integrable h (p) = h (p1) * H (p2) * H (P3) ...

If the only decomposition of P can be written as a part of the prime product I and another part in the previous part of the number of the number of the product J, that is, the exponent of the existence of the mass factor is greater than 1, it added each factor of the μ value is 0, no meaning, only the statistical d becomes the original J times

So at this point h (p) = h (i) * j

Prefix and add it at a later moment.

1#include <bits/stdc++.h>2 3 #defineMoD 1000000094 #defineMaxp 100000005 #defineMAXN 10000000+56 #defineSet (a B) memset (A, (b), sizeof (a))7 #defineFR (i,a,b) for (ll i= (a), _end_= (b); i<=_end_;i++)8 #defineRF (I,b,a) for (ll i= (a), _end_= (b); i>=_end_;i--)9 #defineFe (I,A,B) for (int i=first[(b)],_end_= (a); i!=_end_;i=s[i].next)Ten #defineFEC (I,A,B) for (int &i=cur[(b)],_end_= (a); i!=_end_;i=s[i].next) One  A using namespacestd; -  -typedefLong Longll; the  - ll F[MAXN],H[MAXN]; - ll ans; - intprime[maxn],pri[maxn],tot=0; + intn,m,t; -  + voidRead () A { at #ifndef Online_judge -Freopen ("2693.in","R", stdin); -Freopen ("2693.out","W", stdout); - #endif -   //cin >> T; -scanf"%d",&T); in } -  to voidWrite () + {} -  the voidprint () * { $   //cout << ans << endl;Panax Notoginsengprintf"%lld\n", ans); - } the  + void Get() A { theh[1]=1; +Fr (I,2, MAXP) { -     if(!prime[i]) pri[++tot]=i,h[i]= (i-i*i%mod)%MoD; $     intj=1; $      while(J<=tot && pri[j]*i<=Maxp) { -prime[Pri[j]*i]=1; -       if(i%pri[j]==0 ){ theh[pri[j]*i]=pri[j]*h[i]%MoD; -      Break;Wuyi       } theh[pri[j]*i]=h[pri[j]]*h[i]%MoD; -J + +; Wu     } -   } AboutFr (I,1, MAXP) $F[i]= (f[i-1]+h[i])%MoD; - } -  - ll Sum (ll X,ll y) A { +   return((x+1) *x/2%MOD) * ((y+1) *y/2%MOD)%MoD; the } -  $  the ll Calc (ll X,ll y) the { the   if(x>y) Swap (x, y); thell res=0, i=1, POS; -    while(i<=x) { inPos=min (x/(x/i), y/(y/i)); theRes= (Res+sum (x/i,y/i) * (f[pos]-f[i-1])%mod)%MoD; thei=pos+1; About   } the   return(res+mod)%MoD; the } the  + voidWork () - { the   Get();Bayi    while(t-- ){ the     //cin >> n >> m; thescanf"%d%d",&n,&m); -cn1=Calc (n,m); - print (); the   } the } the  the intMain () - { the read (); the Work (); the write ();94   return 0; the}

Bzoj 2693 Jzptab "The inverse of the Mo"