题目描述
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 
, 
and 
, find the number of integer pairs 
satisfying the following conditions:
,
,
, where 
is the greatest common divisor of 
and 
".
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。
输入输出格式
输入格式：
The first line of the standard input contains one integer 
(
),denoting the number of queries.
The following 
lines contain three integers each: 
, 
and 
(
), separated by single spaces.
Each triplet denotes a single query.
输出格式：
Your programme should write 
lines to the standard output. The 
'th line should contain a single integer: theanswer to the 
'th query from the standard input.
输入输出样例
2 4 5 2 6 4 3
3 2
题解：莫比乌斯反演+分块
其实我写过一边博客上的题跟这个几乎一摸一样，而且这个还不要容斥
在这里偷个懒，贴出题目 [HAOI2011]Problem b
本题卡常数，所以能不用long long就不用
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cstring> 5 using namespace std; 6 typedef long long lol; 7 lol ans; 8 int prime[50001]; 9 lol mu[50001]; 10 int n,m,d,tot; 11 bool vis[50001]; 12 void mobius() 13 {int i,j; 14 mu[1]=1; 15 for (i=2;i<=50000;i++) 16 { 17 if (vis[i]==0) 18 { 19 tot++; 20 prime[tot]=i; 21 mu[i]=-1; 22 } 23 for (j=1;j<=tot,i*prime[j]<=50000;j++) 24 { 25 vis[i*prime[j]]=1; 26 if (i%prime[j]==0) 27 { 28 mu[i*prime[j]]=0; 29 break; 30 } 31 mu[i*prime[j]]=-mu[i]; 32 } 33 } 34 for (i=1;i<=50000;i++) 35 mu[i]+=mu[i-1]; 36 } 37 void solve() 38 {int i; 39 int pos=1; 40 int r=min(n,m); 41 for (i=1;i<=r;i=pos+1) 42 { 43 if (n/(n/i)>m/(m/i)) 44 pos=m/(m/i); 45 else pos=n/(n/i); 46 ans+=(mu[pos]-mu[i-1])*(long long)(n/i)*(long long)(m/i); 47 } 48 } 49 int main() 50 {int T; 51 cin>>T; 52 mobius(); 53 while (T--) 54 { 55 scanf("%d%d%d",&n,&m,&d); 56 n=n/d;m=m/d; 57 ans=0; 58 solve(); 59 printf("%lld ",ans); 60 } 61 }