【NOI2010】能量采集 题解
谨纪念我的第一道手推出来的莫反题。
题目大意:已知 \(n\),\(m\),求 \(\sum\limits_{i=1}^n\sum\limits_{j=1}^m(2\cdot \gcd(i,j)-1)\)。
首先变形一手:
\[\sum\limits_{i=1}^n\sum\limits_{j=1}^m(2\cdot\gcd(i,j)-1)=2\sum\limits_{i=1}^n\sum\limits_{j=1}^m\gcd(i,j)-n\times m \]然后我们只用求出中间那两个 \(\sum\) 就好了。
\[\begin{aligned} \sum\limits_{i=1}^n\sum\limits_{j=1}^m\gcd(i,j)&=\sum\limits_{i=1}^n\sum\limits_{j=1}^m\sum\limits_{d=1}^nd[\gcd(i,j)=d]\\ &=\sum\limits_{d=1}^nd\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}[\gcd(i,j)=1]\\ &=\sum\limits_{d=1}^nd\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}\sum\limits_{x|\gcd(i,j)}\mu(x)\\ &=\sum\limits_{d=1}^nd\sum\limits_{x=1}^{\lfloor\frac{n}{d}\rfloor} \mu(x)\lfloor\frac{n}{dx}\rfloor\lfloor\frac{m}{dx}\rfloor \end{aligned} \]令 \(T=dx\),
\[\begin{aligned} \sum\limits_{d=1}^nd\sum\limits_{x=1}^{\lfloor\frac{n}{d}\rfloor} \mu(x)\lfloor\frac{n}{dx}\rfloor\lfloor\frac{m}{dx}\rfloor &=\sum\limits_{d=1}^nd\sum\limits_{T=1}^n\mu(\frac{T}{d})\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor[d|T]\\ &=\sum\limits_{T=1}^n\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor\sum\limits_{d=1}^nd\cdot\mu(\frac{T}{d})[d|T]\\ &=\sum\limits_{T=1}^n\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor\sum\limits_{d|T}d\cdot\mu(\frac{T}{d}) \end{aligned} \]如何处理后面那个 \(\sum\),考虑狄利克雷卷积。不会的可以看我博客。
因为 \(\varphi*I=Id_1\),又因 \(I*\mu=\epsilon\),所以
\[\varphi=Id_1*\mu \]注意到右边那个 \(\sum\) 其实就是 \(Id_1*\mu\),即 \(\varphi\)。
所以可化为:
\[\sum\limits_{T=1}^n\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor\sum\limits_{d|T}d\cdot\mu(\frac{T}{d})=\sum\limits_{T=1}^n\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor\cdot\varphi(T) \]很明显的整除分块,预处理 \(\varphi\) 的前缀和就好了。
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e5+5;
int n,m;
int cnt,pri[N],phi[N],mu[N],sum[N];
bool flg[N];
void init()
{
mu[1]=1,phi[1]=1;
for(int i=2;i<=N-5;i++)
{
if(!flg[i])
pri[++cnt]=i,phi[i]=i-1,mu[i]=-1;
for(int j=1;j<=cnt&&pri[j]*i<=N-5;j++)
{
flg[i*pri[j]]=1;
if(i%pri[j]==0)
{
phi[i*pri[j]]=phi[i]*pri[j];
break;
}
mu[i*pri[j]]=-mu[i];
phi[i*pri[j]]=phi[i]*phi[pri[j]];
}
}
for(int i=1;i<=N-5;i++)
sum[i]=sum[i-1]+phi[i];
}
int work(int n,int m)
{
if(n>m) swap(n,m);
int res=0;
for(int l=1,r;l<=n;l=r+1)
{
r=min(n/(n/l),m/(m/l));
res+=(sum[r]-sum[l-1])*(n/l)*(m/l);
}
return res;
}
signed main()
{
init();
scanf("%lld%lld",&n,&m);
printf("%lld\n",2*work(n,m)-n*m);
return 0;
}
标签:lfloor,frac,limits,题解,sum,rfloor,mu,采集,NOI2010
From: https://www.cnblogs.com/holmes-wang/p/18212595