题目链接
题解
\[\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)==k]\]
\[=\sum_{i=1}^{? \dfrac{n}{k}?}\sum_{j=1}^{?\dfrac{m}{k}?}[gcd(i,j)==1]\]
利用函数f(x)表示x|(gcd(i,j))中ij的对数,那么原函数:
\[=\sum_{i=1}^{? \dfrac{n}{k}?}\sum_{j=1}^{?\dfrac{m}{k}?}\sum_{d|gcd(i,j)}\mu(d)\]
\[=\sum_{d=1}^{min(\dfrac{n}{k},\dfrac{m}{k})} \mu(d)*\sum_{d|i,i\leq\dfrac{n}{k}}\sum_{d|j,j\leq\dfrac{m}{k}}1\]
\[\sum_{d=1}^{min(?\dfrac{n}{k}?,?\dfrac{m}{k}?)} \mu(d)*?\dfrac{n}{k}?*?\dfrac{m}{k}?)\]
\(?\dfrac{n}{k}?\)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;
inline int read() {
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_wrrj() {
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_wrrj();
int T=read();
for(int a,b,c,d,k;T--;) {
a=read();b=read();c=read();d=read();k=read();
printf("%lld\n",calc(b,d,k)-calc(a-1,d,k)-calc(b,c-1,k)+calc(a-1,c-1,k));
}
return 0;
}