# bzoj 2301 HAOI2011 Problem b

### 题目链接

BZOJ 2301 HAOI2011 Problem b

### 题解

$\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)==k]$

$=\sum_{i=1}^{⌊ \dfrac{n}{k}⌋}\sum_{j=1}^{⌊\dfrac{m}{k}⌋}[gcd(i,j)==1]$

$=\sum_{i=1}^{⌊ \dfrac{n}{k}⌋}\sum_{j=1}^{⌊\dfrac{m}{k}⌋}\sum_{d|gcd(i,j)}\mu(d)$

$=\sum_{d=1}^{min(\dfrac{n}{k},\dfrac{m}{k})} \mu(d)*\sum_{d|i,i\leq\dfrac{n}{k}}\sum_{d|j,j\leq\dfrac{m}{k}}1$

$\sum_{d=1}^{min(⌊\dfrac{n}{k}⌋,⌊\dfrac{m}{k}⌋)} \mu(d)*⌊\dfrac{n}{k}⌋*⌊\dfrac{m}{k}⌋)$

$⌊\dfrac{n}{k}⌋$最多只有$2\sqrt{n}$个取值,预处理$\mu(d)$ $O(\sqrt{n})$回答

### 代码

#include<cstdio>
#include<cstring>
#include<algorithm>
typedef long long LL;
const int maxn=50007;
int x=0;
char c=getchar();
while(c<'0'||c>'9')c=getchar();
while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();
return x;
}
int n,prime[maxn];
bool vis[maxn];long long mu[maxn];
void get_asd() {
mu[1]=1;
for(int i=2;i<=maxn-1;i++) {
if(!vis[i]) prime[++prime[0]]=i,mu[i]=-1;
for(int j=1;j<=prime[0]&&i*prime[j]<=maxn-1;j++) {
vis[i*prime[j]]=1;
if(i%prime[j]==0) {
mu[i*prime[j]]=0;break;
}
mu[i*prime[j]]=-mu[i];
}
}
for(int i=1;i<=maxn-1;i++) mu[i]+=mu[i-1];
}
LL calc(int n,int m,int k) {
n/=k;m/=k;
if(n>m) std::swap(n,m);
LL ans=0;int next=0;
for(int i=1;i<=n;i=next+1) {
next=std::min(n/(n/i),m/(m/i));
ans+=(mu[next]-mu[i-1])*(n/i)*(m/i);
}
return ans;
}
int main() {
get_asd();