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。
1 //It is made by Awson on 2018.1.21 2 #include <set> 3 #include <map> 4 #include <cmath> 5 #include <ctime> 6 #include <queue> 7 #include <stack> 8 #include <cstdio> 9 #include <string> 10 #include <vector> 11 #include <cstdlib> 12 #include <cstring> 13 #include <iostream> 14 #include <algorithm> 15 #define LL long long 16 #define Abs(a) ((a) < 0 ? (-(a)) : (a)) 17 #define Max(a, b) ((a) > (b) ? (a) : (b)) 18 #define Min(a, b) ((a) < (b) ? (a) : (b)) 19 #define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b)) 20 #define writeln(x) (write(x), putchar(‘\n‘)) 21 #define lowbit(x) ((x)&(-(x))) 22 using namespace std; 23 const int N = 50000; 24 void read(int &x) { 25 char ch; bool flag = 0; 26 for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == ‘-‘)) || 1); ch = getchar()); 27 for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar()); 28 x *= 1-2*flag; 29 } 30 void write(LL x) { 31 if (x > 9) write(x/10); 32 putchar(x%10+48); 33 } 34 35 int a, b, k, mu[N+5]; 36 37 void get_mu() { 38 int isprime[N+5], prime[N+5], tot = 0; 39 memset(isprime, 1, sizeof(isprime)); isprime[1] = 0, mu[1] = 1; 40 for (int i = 2; i <= N; i++) { 41 if (isprime[i]) mu[i] = -1, prime[++tot] = i; 42 for (int j = 1; j <= tot && i*prime[j] <= N; j++) { 43 isprime[i*prime[j]] = 0; 44 if (i%prime[j]) mu[i*prime[j]] = -mu[i]; 45 else {mu[i*prime[j]] = 0; break; } 46 } 47 mu[i] += mu[i-1]; 48 } 49 } 50 LL cal(int a, int b) { 51 if (a > b) Swap(a, b); LL ans = 0; 52 for (int i = 1, last; i <= a; i = last+1) { 53 last = Min(a/(a/i), b/(b/i)); 54 ans += (LL)(mu[last]-mu[i-1])*(a/i)*(b/i); 55 } 56 return ans; 57 } 58 void work() { 59 read(a), read(b), read(k); writeln(cal(a/k, b/k)); 60 } 61 int main() { 62 int t; read(t); get_mu(); 63 while (t--) work(); 64 return 0; 65 }
