UOJ221 NOI2016 beauty cycle
tags: number theory, mathematical
The meaning of problems
Allows you to find in \ (K \) the number of values different from each other under pure binary decimals
answer
First question into Italian
\[ Ans=\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=1][\gcd(j,k)=1] \]
Then after a wave of simplification
\[ Ans=\sum_{d=1}^{\min(n,m)}\lfloor\frac nd\rfloor\mu(d)[\gcd(d,k)=1]\sum_{j=1}^{\lfloor\frac md\rfloor}[\gcd(j,k)=1] \]
Then divisible block, just after the half equation calculation, the first half of the equation
\[ \begin{aligned} &\sum_{d=1}^n\mu(d)[\gcd(d,k)=1]\\ =&\sum_{d=1}^n\mu(d)-\sum_{d=1}\mu(d)[\gcd(d,k)>1]\\ =&\sum_{d=1}^n\mu(d)-\sum_{e>1,e|k}\sum_{d=1}^{\lfloor\frac ne\rfloor}\mu(de)[\gcd(de,k)=d]\\ =&\sum_{d=1}^n\mu(d)-\sum_{e>1,e|k}\sum_{d=1}^{\lfloor\frac ne\rfloor}\mu(d)\mu(e)[\gcd(d,e)=1][\gcd(de,k)=e]\\ =&\sum_{d=1}^n\mu(d)-\sum_{e>1,e|k}\mu(e)\sum_{d=1}^{\lfloor\frac ne\rfloor}\mu(d)[\gcd(d,k)=1] \end{aligned} \]
Du then vigorously enough to teach sieve
Half an expression not simply push out, I really good food
#include<map>
#include<cstdio>
#include<vector>
#include<unordered_map>
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
template<typename T>inline void rd(T&x){int fl=0,ch;while(ch=getchar(),ch<48||57<ch)fl^=!(ch^45);x=(ch&15);while(ch=getchar(),47<ch&&ch<58)x=x*10+(ch&15);if(fl)x=-x;}
template<typename T>inline void pt(T x){if(x<0)putchar('-'),x=-x;if(x>9)pt(x/10);putchar(x%10+48);}
template<typename T>inline void pt(T x,int ch){pt(x),putchar(ch);}
template<typename T>inline T max(const T&x,const T&y){return x<y?y:x;}
template<typename T>inline T min(const T&x,const T&y){return x<y?x:y;}
int gcd(int a,int b){while(a^=b^=a^=b%=a,a);return b;}
const int N=1000005;
int n,m,k,vis[N],p[N],mu[N],s[N],s1[N];std::vector<int>v;std::unordered_map<int,int>mmp,mmp1;
void pre(){
vis[0]=vis[1]=1,mu[1]=1;
for(int i=2;i<N;++i){
if(!vis[i])p[++*p]=i,mu[i]=-1;
for(int j=1;j<=*p&&i*p[j]<N;++j){
vis[i*p[j]]=1;
if(i%p[j])mu[i*p[j]]=-mu[i];
else break;
}
}
for(int i=1;i<N;++i)s[i]=s[i-1]+mu[i];
for(int i=1;i<N;++i)s1[i]=s1[i-1]+mu[i]*(gcd(i,k)==1);
}
int F0(int n){
if(n<N)return s[n];
if(mmp.count(n))return mmp[n];
int res=1;
for(int l=2,r;l<=n;l=r+1){
r=n/(n/l);
res-=F0(n/l)*(r-l+1);
}
return mmp[n]=res;
}
int F1(int n){
if(n<N)return s1[n];
if(mmp1.count(n))return mmp1[n];
int res=F0(n);
for(int i=1;i<(int)v.size();++i){
int e=v[i];
res-=mu[e]*F1(n/e);
}
return mmp1[n]=res;
}
int F2(int n){
int res=0;
for(int i=0;i<(int)v.size();++i){
int d=v[i];
res+=mu[d]*(n/d);
}
return res;
}
signed main(){
rd(n),rd(m),rd(k);pre();
for(int i=1;i<=k;++i)if(mu[i]&&k%i==0)v.push_back(i);
long long ans=0;
for(int l=1,r;l<=n&&l<=m;l=r+1){
r=min(m/(m/l),n/(n/l));
ans+=(long long)n/l*(F1(r)-F1(l-1))*F2(m/l);
}
pt(ans,'\n');
return 0;
}