[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 ;
 }