Bzoj 1101 Zap (Möbius inversion)

Http://www.lydsy.com/JudgeOnline/problem.php?id=1101

Given a,b,d, how many gcd (x, y) ==d (1<=x<=a&&1<=y<=b) are asked

Ideas:

ΣGCD (x, y) ==d (1<=x<=a,1<=y<=b)

=

ΣGCD (x, y) ==1 (1<=x<=a/d,1<=y<=b/d)

Make g (i) =num (I|GCD (x, y)) =n/i*m/i

G (i) =num (I=GCD (x, y))

G (i) =σg (d) *u (d/i) (i|d)

The answer is G (1)

G (1) =σg (i) *u (i) (1<=i<=min (n,m))

=σ (n/i) * (m/i) *u (i)

So make a prefix of U (i) and so that you can work together n/i and m/i the same I

1#include <algorithm>2#include <cstdio>3#include <cmath>4#include <cstring>5#include <iostream>6 intmul[200005],mark[200005],sum[200005],p[200005];7 intRead () {8     CharCh=getchar ();intt=0, f=1;9      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();}Ten      while('0'<=ch&&ch<='9') {t=t*Ten+ch-'0'; ch=GetChar ();} One     returnt*F; A } - voidinit () { -mul[1]=1; the      for(intI=2; i<=50000; i++){ -         if(!Mark[i]) { -p[++p[0]]=i; -mul[i]=-1; +         } -          for(intj=1; j<=p[0]&&i*p[j]<=50000; j + +){ +mark[p[j]*i]=1; A             if(I%p[j]) mul[p[j]*i]=mul[i]* (-1); at             Else { -mul[p[j]*i]=0; -                  Break; -             } -         } -     } insum[0]=0; -      for(intI=1; i<=50000; i++) tosum[i]=sum[i-1]+Mul[i]; + } - intCalintAintb) { the     if(a>b) Std::swap (a); *     intres=0, n=a,m=b; $      for(intI=1, j;i<=a;i=j+1){Panax NotoginsengJ=std::min (n/(n/i), m/(m/i)); -res+= (a/i) * (b/i) * (sum[j]-sum[i-1]); the     } +     returnRes; A } the intMain () { +     intq=read (); - init (); $      while(q--){ $         intA=read (), B=read (), d=read (); -printf"%d\n", Cal (a/d,b/d)); -     } the}

Bzoj 1101 Zap (Möbius inversion)