"Bzoj" 3309:dzy Loves Math-mo-colorimetric inversion optimization

3309:dzy Loves MathDescription

For positive integer n, define the maximum power exponent of the mass factor of N for f (n). For example F (1960) =f (2^3 * 5^1 * 7^2) =3, F (10007) =1, f (1) = 0.
Given a positive integer, a, B, for Sigma (Sigma (GCD (I,J))) (I=1..A, j=1..b).

Input

The first line is a number T, which indicates the number of queries.
Next T line, two number per line, a, B, indicates a query.

Output

For each query, output a non-negative integer as the answer.

Sample Input4
7558588 9653114
6514903 4451211
7425644 1189442
6335198 4957
Sample Output35793453939901
14225956593420
4332838845846
15400094813HINT

"Data Size"

t<=10000

1<=a,b<=10^7

Reference post: POPOQQQ

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I am mainly talking about why ( -1) The contribution of k+1 is only when all the power exponents are equal when the t factor decomposition;

In order to optimize the inner loop, we need to put a A, B, pretreatment, then it becomes the third step, the T-factor decomposition to T = P1^a1*p2^a2*...*pk^ak;

Möbius inversion formula is easy to know, d| t want to make MU[T/D]! = 0 only the coefficients of each factor are <= 1, then the total D is 2k (K is the number of T-factor);

<1> When a1 = ... = AK = a , for F[d] easy to know only when all take is A-1, only for A-1, then there is only one situation; contribution is (A-1) * ( -1) ^k = A * ( -1) ^k + ( -1) ^k;

F[D] = a need to be solved by combining number, (the number of combinations is C (k,i), where I is the number of A-1, i.e. i = 0,1,..., k-1), here i! = k; (i = k is f[d] = A-1),

The contribution at this time is a * (( -1) ^0*c (k,0) + ( -1) ^1*c (k,1) +...+ ( -1) ^ (k-1) *c (k,k-1)) =?

Look directly (1-1) ^k will find the missing A * ( -1^k*c (k,k)) = a * ( -1) ^k, so after the first case, need-( -1) ^k = ( -1) ^k+1

<2> when there is ai! = AJ, the maximum power goes to a or A-1 is f[d], so it is impossible to have a sub-case discussion, while the other goes to AJ | Aj-1 will cause the odd and even change and make the f[d] become 0;

So the conclusion is that only a1 = A2 = ... = AK = A, σf[d]*μ (t/d) is not 0, and f[d] is not required to calculate, to get only the odd parity of k = > ( -1) ^ (k+1)

The preprocessing time complexity is the same as the Sieve method O (n), then the data processing time of each group is O (sqrt (n)), which is the block acceleration;

Coding Details: It uses recursion to get g[]: The current value is 0 or 1, i.e. ( -1) ^ (k+1), followed by prefix and is for block acceleration;

The same recursion is also used in the A[i]: the power exponent of the minimum quality factor of I, the exponential product of the VAL[I]:I minimum mass factor (P[1]^a[i], which is to judge A[i] when the g[i] is obtained) [A[j/val[i]] That is, whether the power exponent of the smallest mass factor is equal to the second lowest power exponent. Note that this is a recursive relationship, that is, does not mean to compare the smallest two power exponent, which is equivalent to the Markov chain ~ ~ ~ (just see someone else's code understanding) ingenious Orz

#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<stdlib.h>using namespacestd;#defineRep0 (I,L,R) for (int i = (l); i < (R); i++)#defineREP1 (I,L,R) for (int i = (l); I <= (r); i++)#defineRep_0 (i,r,l) for (int i = (r); i > (l); i--)#defineRep_1 (i,r,l) for (int i = (r); I >= (l); i--)#defineINF 0x7ffffffftypedefLong Longll;template<typename t>voidRead1 (T &m) {T x=0, f=1;CharCh=GetChar ();  while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();}  while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} M= x*F;} Template<typename t>voidRead2 (T &a,t &b) {Read1 (a); Read1 (b);} Template<typename t>voidRead3 (T &a,t &b,t &c) {Read1 (a); Read1 (b); Read1 (c);}Const intN = 1e7+7;intVis[n],g[n],p[n],a[n],val[n];voidinit () {rep0 (I,2, N) {        if(Vis[i] = =0) {p[++p[0]] =i; G[i]=1; A[i]=1; Val[i]=i; }         for(intj =1;p [J]*i < n;j++) {vis[i*P[J]] =1; if(I%p[j] = =0) {A[i*P[J]] = a[i]+1;//the power exponent of the minimum factorization;VAL[I*P[J]] = val[i]*P[j]; intTMP = i/Val[i]; if(TMP = =1) G[i*p[j]] =1;//there's only one quality factor .                ElseG[i*P[J]] = (a[tmp] = = A[i*p[j]]?-g[tmp]:0);  Break; } A[p[j]*i] =1;//indicates the number of p[j] is 1;VAL[I*P[J]] =P[j]; G[P[J]*i] = (a[i] = =1?-g[i]:0); }} rep0 (I,1, N) g[i] + = g[i-1];}intMain () {init (); intn,m,t,i,j;    Read1 (T);  while(t--) {read2 (n,m); ll ans=0; if(N >m) swap (N,M);  for(i =1; I <= n;i = j +1) {J= Min (n/(n/i), m/(m/i)); Ans+ = 1ll* (G[j]-g[i-1]) * (n/i) * (m/i); } printf ("%lld\n", ans);//I64D is the result of WA, which runs the system differently. }    return 0;}

"Bzoj" 3309:dzy Loves Math-mo-colorimetric inversion optimization