Bzoj 2301 "HAOI2011" Problem B

Description for the given n queries, how many pairs (x, y) are asked each time, satisfy A≤x≤b,c≤y≤d, and gcd (x, y) = K,GCD (x, y) functions are greatest common divisor of x and Y. Input

The first line is an integer n, and the next n rows are five integers per line, representing A, B, C, D, K, respectively.

Output

A total of n rows, one integer per line representing the number of pairs (x, y) that satisfy the requirement

HINT

100% of the data meet: 1≤n≤50000,1≤a≤b≤50000,1≤c≤d≤50000,1≤k≤50000

After so long finally wrote the Möbius inversion of the introductory question tat ...

This problem mainly uses a property of the Möbius function, for any positive integer $n$, there are: $$\sum_{d|n} \mu (d) = \left \{\begin{aligned} &1 & (N=1) \ &0 & (n>1 ) \end{aligned} \right.$$

So $[GCD (i,j) =1]$ This expression can be expressed as: $$\SUM_{D|GCD (I,J)}\mu (d) $$

So Möbius inversion is especially useful for dealing with problems such as $GCD (x, y) =1$.

Möbius inversion stamp here

Refer to the Huang long blog (I have turned to the previous page ...). If you want to see the code, please turn to the next article)

After all, I just looked at Huang long blog made out of ...

Paste the following code:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include < Cmath> #define FILE (s) freopen (S ".", "R", stdin), Freopen (S ". Out", "w", stdout) #define MAXN 50010using namespace std; typedef long LONG Llg;int t,a,b,c,d,k,ls;int mu[maxn],s[maxn],w[maxn];bool vis[maxn];int getint () {int W=0;bool Q=0;char C=getchar (); while (c> ' 9 ' | | c< ' 0 ') &&c!= '-') C=getchar (), if (c== '-') C=getchar (), Q=1;while (c>= ' 0 ' &&c<= ' 9 ') w=w*10+c-' 0 ', C=getchar (); return q?-w:w;} void Get () {//Linear sieve primes with the mu[1]=1;for function (int i=2;i<maxn;i++) {if (!vis[i]) s[++ls]=i,mu[i]=-1;for (int j=1;j<=ls && s[j]*i<maxn;j++) {vis[s[j]*i]=1;if (i%s[j]) Mu[s[j]*i]=-mu[i];else{mu[s[j]*i]=0;break;}}} for (int i=1;i<maxn;i++) w[i]=w[i-1]+mu[i];} LLG F (int n,int m) {//Find x in [1,n], y in [1,m] answer LLG ans=0;if (n>m) swap (n,m); for (int i=1,nt;i<=n;i=nt+1) {nt=min (n/(n /i), m/(m/i)), ans+= (LLG) (W[nt]-w[i-1]) * (LLG) (n/i) * (LLG) (m/i);} return ans;} int main () {File ("a"); get (); T=getint (); while (t--) {a=getint (); B=getint (); C=getint ();d =getint (); K=getint (); a--; c--;//note border a/=k; b/=k; c/=k; d/=k;printf (" %lld\n ", F (b,d)-F (a,d)-F (b,c) +f (a,c));//conversion to prefix and repulsion solution}return 0;}

　　

Bzoj 2301 "HAOI2011" Problem B