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 aa , bb and dd , 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 xx and yy ".
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 \(n\) (\(1\le n\le 50\ 000\) ),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.
输出格式：
Your programme should write nn lines to the standard output. The \(i\) ‘th line should contain a single integer: theanswer to the \(i\) ‘th query from the standard input.
输入输出样例
输入样例#1：
2
4 5 2
6 4 3
输出样例#1：
3
2
题解
这题其实还是上一篇的弱化版，f(x)和F(x)的定义一样，最后的式子是：
\[
ans=f(d)=\sum_{T=1}^{min(a,b)}\mu(\frac{T}{n})\lfloor \frac{a}{T} \rfloor \lfloor \frac{b}{T} \rfloor
\]
一样的整除分块，一样的前缀和
#include<bits/stdc++.h>
#define ll long long
const int MAXN=50000+10;
ll T,mu[MAXN],s[MAXN],prime[MAXN],cnt;
bool vis[MAXN];
template<typename T> inline void read(T &x)
{
T data=0,w=1;
char ch=0;
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')w=-1,ch=getchar();
while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
x=data*w;
}
template<typename T> inline void write(T x,char c='\0')
{
if(x<0)putchar('-'),x=-x;
if(x>9)write(x/10);
putchar(x%10+'0');
if(c!='\0')putchar(c);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
inline void init()
{
memset(vis,1,sizeof(vis));
vis[0]=vis[1]=0;
mu[1]=1;
for(register int i=2;i<MAXN;++i)
{
if(vis[i])
{
prime[++cnt]=i;
mu[i]=-1;
}
for(register int j=1;j<=cnt&&i*prime[j]<MAXN;++j)
{
vis[i*prime[j]]=0;
if(i%prime[j])mu[i*prime[j]]=-mu[i];
else break;
}
}
for(register int i=1;i<MAXN;++i)s[i]=s[i-1]+mu[i];
}
inline ll solve(ll a,ll b,ll d)
{
ll res=0;
for(register ll i=1;;)
{
if(i>min(a,b))break;
ll j=min(a/(a/i),b/(b/i));
res+=(a/i)*(b/i)*(s[j/d]-s[(i-1)/d]);
i=j+1;
}
return res;
}
int main()
{
init();
read(T);
while(T--)
{
ll a,b,d;
read(a);read(b);read(d);
write(solve(a,b,d),'\n');
}
return 0;
}