标签：void const return math bre ios nbsp else pre
题意：$\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^m {lcm(i,j)} } $
解题关键：
$\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^m {lcm(i,j)} } = \sum\limits_{i = 1}^n {\sum\limits_{j = 1}^m {\frac{{i*j}}{{\gcd (i,j)}}} } $
枚举gcd，上式化为：
$\sum\limits_{d = 1}^{\min (n,m)} {d\sum\limits_{\begin{array}{*{20}{c}}
{\gcd (i,j) = = 1}\\
{1 < = i < = n/d}\\
{1 < = j < = m/d}
\end{array}} {i*j} } $
令
$f(n,m,k) = \sum\limits_{\begin{array}{*{20}{c}}
{\gcd (i,j) = = k}\\
{1 < = i < = n}\\
{1 < = j < = m}
\end{array}} {i*j} $
由于
$sum(n,m) = \sum\limits_{\begin{array}{*{20}{c}}
{1 < = i < = n}\\
{1 < = j < = m}
\end{array}} {i*j} = \frac{{n(n + 1)}}{2}\frac{{m(m + 1)}}{2}$
$\begin{array}{l}
F(n,m,k) = \sum\limits_{k|d} {f(n,m,d) = \sum\limits_{\begin{array}{*{20}{c}}
{1 < = i < = n}\\
{1 < = j < = m}\\
{k|\gcd (i,j)}
\end{array}} {i*j} = {k^2}sum(\left\lfloor {\frac{n}{k}} \right\rfloor ,\left\lfloor {\frac{m}{k}} \right\rfloor )} \\
f(n,m,k) = \sum\limits_{k|d} {u(\frac{d}{k})F(n,m,d)}
\end{array}$
而此题中，$k==1$,
则，
$\begin{array}{l}
f(n,m,1) = \sum\limits_{d = 1}^{\min (n,m)} {u(d)F(n,m,d)} \\
= \sum\limits_{d = 1}^{\min (n,m)} {u(d){d^2}sum(\left\lfloor {\frac{n}{d}} \right\rfloor ,\left\lfloor {\frac{m}{d}} \right\rfloor )} \\
ans = \sum\limits_{d = 1}^{\min (n,m)} {d*f(\left\lfloor {\frac{n}{d}} \right\rfloor ,\left\lfloor {\frac{m}{d}} \right\rfloor ,1)}
\end{array}$
求解ans和$f$函数复杂度都是$O(\sqrt n )$，所以最终复杂度为$O(n)$。
1 #include<cstdio> 2 #include<cstring> 3 #include<cstdlib> 4 #include<algorithm> 5 #include<cmath> 6 #include<iostream> 7 #define Sum(x,y) (x*(x+1)/2%mod*(y*(y+1)/2%mod)%mod) 8 using namespace std; 9 typedef long long ll; 10 const ll mod=20101009; 11 const ll maxn=10000000+200; 12 ll n,m; 13 bool vis[maxn]; 14 ll prime[maxn],mu[maxn],sum1[maxn]; 15 void init_mu(ll n){ 16 ll cnt=0; 17 mu[1]=1; 18 for(ll i=2;i<n;i++){ 19 if(!vis[i]){ 20 prime[cnt++]=i; 21 mu[i]=-1; 22 } 23 for(ll j=0;j<cnt&&i*prime[j]<n;j++){ 24 vis[i*prime[j]]=1; 25 if(i%prime[j]==0){mu[i*prime[j]]=0;break;} 26 else { mu[i*prime[j]]=-mu[i];} 27 } 28 } 29 sum1[0]=0; 30 for(ll i=1;i<n;i++) sum1[i]=(sum1[i-1]+1ll*mu[i]*i*i)%mod; 31 } 32 ll calf(ll n,ll m){ 33 ll ans=0,pos,len=min(n,m); 34 for(ll i=1;i<=len;i=pos+1){ 35 pos=min(n/(n/i),m/(m/i)); 36 //ans+=(sum1[pos]-sum1[i-1])%mod*((n/i)*((n/i)+1)/2%mod*(((m/i)+1)*(m/i)/2%mod)%mod); 37 ans+=(sum1[pos]-sum1[i-1])%mod*Sum(n/i,m/i)%mod;//最好用函数写出 38 ans%=mod; 39 40 } 41 return ans; 42 } 43 ll calres(ll n,ll m){ 44 ll ans=0,pos,len=min(n,m); 45 for(ll i=1;i<=len;i=pos+1){ 46 pos=min(n/(n/i),m/(m/i)); 47 ans+=(i+pos)*(pos-i+1)/2%mod*calf(n/i,m/i)%mod; 48 ans%=mod; 49 } 50 return (ans%mod+mod)%mod; 51 } 52 53 int main(){ 54 scanf("%lld%lld",&n,&m); 55 init_mu(min(n,m)+10); 56 printf("%lld\n",calres(n,m)); 57 return 0; 58 }
[bzoj2154]Crash的数字表格(mobius反演)
标签：void const return math bre ios nbsp else pre
原文地址：http://www.cnblogs.com/elpsycongroo/p/7624268.html
