[POI 2007]ZAP-Queries

Description

Byteasar the Cryptographer works on breaking the code of BSA (Byteotian Security Agency). He has alreadyfound out that whilst deciphering a message he will have to answer multiple queries of the form"for givenintegers $a$, $b$ and $d$, find the number of integer pairs $(x,y)$ satisfying the following conditions:

$1\le x\le a$,$1\le y\le b$,$gcd(x,y)=d$, where $gcd(x,y)$ is the greatest common divisor of $x$ and $y$".

Byteasar would like to automate his work, so he has asked for your help.

TaskWrite a programme which:

reads from the standard input a list of queries, which the Byteasar has to give answer to, calculates answers to the queries, writes the outcome to the standard output.

FGD正在破解一段密码，他需要回答很多类似的问题：对于给定的整数a,b和d，有多少正整数对x,y，满足x<=a，y<=b，并且gcd(x,y)=d。作为FGD的同学，FGD希望得到你的帮助。

Input

The first line of the standard input contains one integer $n$ ($1\le n\le 50\ 000$),denoting the number of queries.

The following $n$ lines contain three integers each: $a$, $b$ and $d$($1\le d\le a,b\le 50\ 000$), separated by single spaces.

Each triplet denotes a single query.

Output

Your programme should write $n$ lines to the standard output. The $i$'th line should contain a single integer: theanswer to the $i$'th query from the standard input.

Sample Input

2
4 5 2
6 4 3

Sample Output

3
2

题解

双倍经验。去掉容斥就好了，分析见超链接。

 //It is made by Awson on 2018.1.21

 #include <set>

 #include <map>

 #include <cmath>

 #include <ctime>

 #include <queue>

 #include <stack>

 #include <cstdio>

 #include <string>

 #include <vector>

 #include <cstdlib>

 #include <cstring>

 #include <iostream>

 #include <algorithm>

 #define LL long long

 #define Abs(a) ((a) < 0 ? (-(a)) : (a))

 #define Max(a, b) ((a) > (b) ? (a) : (b))

 #define Min(a, b) ((a) < (b) ? (a) : (b))

 #define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))

 #define writeln(x) (write(x), putchar('\n'))

 #define lowbit(x) ((x)&(-(x)))

 using namespace std;

 const int N = ;

 void read(int &x) {

     char ch; bool flag = ;

     for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || ); ch = getchar());

     for (x = ; isdigit(ch); x = (x<<)+(x<<)+ch-, ch = getchar());

     x *= -*flag;

 }

 void write(LL x) {

     if (x > ) write(x/);

     putchar(x%+);

 }

 int a, b, k, mu[N+];

 void get_mu() {

     int isprime[N+], prime[N+], tot = ;

     memset(isprime, , sizeof(isprime)); isprime[] = , mu[] = ;

     for (int i = ; i <= N; i++) {

     if (isprime[i]) mu[i] = -, prime[++tot] = i;

     for (int j = ; j <= tot && i*prime[j] <= N; j++) {

         isprime[i*prime[j]] = ;

         if (i%prime[j]) mu[i*prime[j]] = -mu[i];

         else {mu[i*prime[j]] = ; break; }

     }

     mu[i] += mu[i-];

     }

 }

 LL cal(int a, int b) {

     if (a > b) Swap(a, b); LL ans = ;

     for (int i = , last; i <= a; i = last+) {

     last = Min(a/(a/i), b/(b/i));

     ans += (LL)(mu[last]-mu[i-])*(a/i)*(b/i);

     }

     return ans;

 }

 void work() {

     read(a), read(b), read(k); writeln(cal(a/k, b/k));

 }

 int main() {

     int t; read(t); get_mu();

     while (t--) work();

     return ;

 }

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