机房最后一个学懵逼钨丝的人
题目一
链接
题目没找到
求\(\sum_{1}^{n}\sum_{1}^{m}gcd(i,j)\)
式子
\(\sum\limits_{i=1}^{N}\sum\limits_{j=1}^{M} gcd(i,j)\)
\(\sum\limits_{k=1}^{min(N,M)} k*\sum\limits_{i=1}^{N}\sum\limits_{j=1}^{M} [gcd(i,j)==k]=\sum\limits_{k=1}^{min(N,M)} k*\sum\limits_{i=1}^{\frac{N}{k}}\sum\limits_{j=1}^{\frac{M}{k}} [gcd(i,j)==1]\)
\(看这部分\sum\limits_{i=1}^{N}\sum\limits_{j=1}^{M} [gcd(i,j)==k]\)
设\(g(d)=\sum\limits_{i=1}^{N}\sum\limits_{j=1}^{M} [gcd(i,j)==d]\)
\(f(n)=\sum\limits_{n|d}g(d)=[\frac{N}{d}][\frac{M}{d}]\)（显然，不说了）
\(g(n)=\sum\limits_{n|d}\mu(\frac{d}{n})f(d)\)
\(g(1)=\sum\limits_{i=1}^{min(N,M)}\mu(i)[\frac{N}{i}][\frac{M}{i}]\)
\(ans=\sum\limits_{k=1}^{min(N,M)} k*\sum\limits_{i=1}^{min(N/k,M/k)}\mu(i)[\frac{N/k}{i}][\frac{M/k}{i}]\)
除法分块嵌套复杂度\(O（n）\)
注意
除数不为0
代码
/*
segima gcd n,m
*/
#include <algorithm>
#include <iostream>
#include <cstdio>
#define ll long long
using namespace std;
const int N=15000007,mod=1e9+7;
int read() {
int x=0,f=1;char s=getchar();
for(;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1;
for(;s>='0'&&s<='9';s=getchar()) x=x*10+s-'0';
return x*f;
}
int pri[N/10],mu[N],cnt;
bool vis[N];
void Euler(int n) {
vis[1]=mu[1]=1;
for(int i=2;i<=n;++i) {
if(!vis[i]) pri[++cnt]=i,mu[i]=-1;
for(int j=1;j<=cnt&&i*pri[j]<=n;++j) {
vis[ i*pri[j] ] = 1;
if(i % pri[j] == 0) {
mu[ i*pri[j] ] = 0;
break;
} else
mu[ i*pri[j] ] = -mu[i];
}
}
for(int i=1;i<=n;++i) mu[i]+=mu[i-1];
}
int g(int N,int M,int n) {
int ans=0;
for(int l=1,r=1;l<=n;l=r+1) {
r=min(N/(N/l),M/(M/l));
ans=(ans+1LL*(mu[r]-mu[l-1]+mod)%mod*(N/l)%mod*(M/l)%mod)%mod;
}
return ans;
}
ll calc(int a) {return 1LL*a*(a+1)%mod*500000004%mod;}
int main() {
int N=read(),M=read(),n=min(N,M),ans=0;
Euler(n);
for(int l=1,r=1;l<=n;l=r+1) {
r=min(N/(N/l),M/(M/l));
ans=(1LL*ans+(1LL*(calc(r)-calc(l-1))%mod+mod)%mod
*1LL*g(N/l,M/l,min(N/l,M/l)))%mod;
}
cout<<ans<<"\n";
// ans=0;
// for(int i=1;i<=N;++i) {
// for(int j=1;j<=M;++j) {
// ans=(ans+__gcd(i,j))%mod;
// }
// }
// cout<<ans;
return 0;
}
题目
类似的 差不多=多组询问的T1
https://www.luogu.org/problemnew/show/P2257
链接&&问题
\(\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}[gcd(i,j)==pri]\)
公式
\(\sum\limits_{k=1}^{N}\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}[gcd(i,j)==k]\)
扣出\(\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}[gcd(i,j)==k]\)
啊哈，熟悉的套路
$ 此处省略，在T1或T2都有\(
\)ans=\sum\limits_{p=pri}^{min(N,M)} \sum\limits_{i=1}^{N/p}\mu(i)[\frac{N/p}{i}][\frac{M/p}{i}]\(
A了？1000组询问T飞了
\)\sum\limits_{p=pri} \sum\limits_{i=1}^{N/p}\mu(i)[\frac{N/p}{i}][\frac{M/p}{i}]\(
\)\sum\limits_{p=pri} \sum\limits_{i=1}^{N/p}\mu(\frac{k}{p})[\frac{N}{k}][\frac{M}{k}]\(
\)\sum\limits_{k=1}^{N}\sum\limits_{p=pri,p|k} \mu(\frac{k}{p})[\frac{N}{k}][\frac{M}{k}]\(
\)z(k)=\sum\limits_{p=pri,p|k} \mu(\frac{k}{p})$
线性筛预处理出z的前缀和就可以了
它的证明不错
https://orzsiyuan.com/articles/problem-Luogu-2257-YY-GCD/
代码
#include <algorithm>
#include <iostream>
#include <cstdio>
#define ll long long
using namespace std;
const int N=10000007,mod=1e9+7;
int read() {
int x=0,f=1;char s=getchar();
for(;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1;
for(;s>='0'&&s<='9';s=getchar()) x=x*10+s-'0';
return x*f;
}
int pri[N/10],mu[N],cnt,z[N];
bool vis[N];
void Euler(int n) {
vis[1]=mu[1]=1;
for(int i=2;i<=n;++i) {
if(!vis[i]) pri[++cnt]=i,mu[i]=-1,z[i]=mu[1];
for(int j=1;j<=cnt&&i*pri[j]<=n;++j) {
vis[ i*pri[j] ] = 1;
if(i % pri[j] == 0) {
mu[ i*pri[j] ] = 0;
z[ i*pri[j] ] = mu[i];
break;
} else {
mu[ i*pri[j] ] = -mu[i];
z[ i*pri[j] ] = -z[i] + mu[i];
}
}
}
for(int i=1;i<=n;++i) z[i]+=z[i-1];
}
int g(int N,int M,int n) {
ll ans=0;
for(int l=1,r=1;l<=n;l=r+1) {
r=min(N/(N/l),M/(M/l));
ans=ans+1LL*(z[r]-z[l-1])*(N/l)*(M/l);
}
return ans;
}
int main() {
Euler(10000000);
int T=read();
while(T--) {
int a=read(),b=read();
printf("%lld\n",g(a,b,min(a,b)));
}
return 0;
}
bzoj1101
链接
https://www.lydsy.com/JudgeOnline/problem.php?id=1101
式子
同楼上，或许，比他还要简单的多
代码
#include <algorithm>
#include <iostream>
#include <cstdio>
#define ll long long
using namespace std;
const int N=100007,mod=1e9+7;
int read() {
int x=0,f=1;char s=getchar();
for(;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1;
for(;s>='0'&&s<='9';s=getchar()) x=x*10+s-'0';
return x*f;
}
int pri[N/10],mu[N],cnt;
bool vis[N];
void Euler(int n) {
vis[1]=mu[1]=1;
for(int i=2;i<=n;++i) {
if(!vis[i]) pri[++cnt]=i,mu[i]=-1;
for(int j=1;j<=cnt&&i*pri[j]<=n;++j) {
vis[ i*pri[j] ] = 1;
if(i % pri[j] == 0) {
mu[ i*pri[j] ] = 0;
break;
} else
mu[ i*pri[j] ] = -mu[i];
}
}
for(int i=1;i<=n;++i) mu[i]+=mu[i-1];
}
int g(int N,int M,int n) {
ll ans=0;
for(int l=1,r=1;l<=n;l=r+1) {
r=min(N/(N/l),M/(M/l));
ans=ans+1LL*(mu[r]-mu[l-1])*(N/l)*(M/l);
}
return ans;
}
int main() {
// freopen("11.in","r",stdin);
// freopen("a.out","w",stdout);
Euler(100000);
int T=read();
while(T--) {
int a=read(),b=read(),c=read();
printf("%lld\n",g(a/c,b/c,min(a/c,b/c)));
}
return 0;
}