Bzoj 1101 ([poi2007]zap-satisfies the number of X<=A&&Y<=B&&GCD (x, y) =d)

1101: [Poi2007]zap time limit: ten Sec Memory Limit: 162 MB
Submit: 1646 Solved: 577
[Submit] [Status] [Discuss] Description

FGD is cracking a cipher, and he needs to answer a number of similar questions: for a given integer, a, B, and D, how many positive integers are to X, Y, X<=a,y<=b, and gcd (x, y) =d. As a classmate of FGD, FGD hopes to get your help.

Input

The first line contains a positive integer n, which indicates that there are altogether n groups of queries. (1<=n<= 50000) next n rows, each line represents a query, three positive integers per line, respectively, A,b,d. (1<=d<=a,b<=50000)

Output

For each set of queries, output to the output file zap.out a positive integer that represents the integer logarithm that satisfies the condition.

Sample Input2
4 5 2
6 4 3Sample Output3
2HINT

For the first set of queries, an integer pair that satisfies the condition has (2,2), (2,4), (4,2). For the second set of queries, an integer pair that satisfies the condition has (6,3), (3,3).

Source

This is entitled Mobius Inversion:

PS: The calculation process cannot use long long or tle

Formula derivation

Let's take a look at popoqqq ppt "Möbius inversion"

#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include < functional> #include <iostream> #include <cmath>using namespace std; #define for (i,n) for (int i=1;i<=n ; i++) #define Fork (i,k,n) for (int i=k;i<=n;i++) #define MEM (a) memset (A,0,sizeof (a)); #define MAXN (50000+10) typedef Long Long ll;int p[maxn]={0},tot;bool b[maxn]={0};int mu[maxn]={0},sum[maxn]={0};inline int read () {int X=0,f=1;char ch= GetChar (); while (ch< ' 0 ' &&ch> ' 9 ') {if (ch== '-') f=-1; Ch=getchar ();} while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 '; Ch=getchar ();} return x*f;} void Make_prime (int n) {tot=0; mu[1]=1; Fork (i,2,n) {if (!b[i]) p[++tot]=i,mu[i]=-1; for (J,tot) {if (i*p[j]>n) break;b[i*p[j]]=1;if (i%p[j]==0) {mu[i*p[j]]=0;  Mu[i*p[j]]=-mu[i]; }}sum[0]=0; for (I,n) sum[i]=sum[i-1]+mu[i];} int N,m,d;int calc () {int ans=0;for (int i=1,last=1;i<=n;i=last+1) {last=min (n/(n/i), m/(m/i)), ans+= (sum[last]-sum[ I-1]) * (n/i) * (m/i);} printf ("%d\n", aNS); return ans;} int main () {//freopen ("bzoj1101.in", "R", stdin); MEM (P) mem (b) Mem (MU) mem (sum) int N = 50000;make_prime (n); int T;  T=read (); while (t--) {n=read (); M=read (); D=read (); if (n>m) swap (n,m); N/=d,m/=d;calc ();} return 0;}

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Bzoj 1101 ([poi2007]zap-satisfies the number of X<=A&&Y<=B&&GCD (x, y) =d)