BZOJ2226:[SPOJ5971]LCMSum

Description

Given n, calculate the sum LCM(1,n) + LCM(2,n) + .. + LCM(n,n), where LCM(i,n) denotes the Least Common Multiple of the integers i and n.

Input

The first line contains T the number of test cases. Each of the next T lines contain an integer n.

Output

Output T lines, one for each test case, containing the required sum.

Sample Input

3
1
2
5

Sample Output

1
4
55

HINT

1 <= T <= 300000
1 <= n <= 1000000

题解：

题意即求∑LCM(i,n)（1<=i<=n）。

枚举gcd，统计对答案的贡献。

原本我采用的方法是容斥，求出n的因数表后，由大到小枚举gcd[i]，并把更小的gcd[j]的贡献减去相应的值。

复杂度还可以，但是常数非常大，在BZ上过不了。

有一个常数更小的方法：枚举gcd后，我们需要知道1~n div gcd-1中所有与n div gcd互质的数的和。

设m=n div gcd。若x与m互质，则m-x与m互质，即与m互质的数成对出现，所以与m互质的数的和为m*φ(m)div 2。（m<=2时依旧成立）

线性筛预处理出欧拉函数，就可以快速求值了。

代码：

TLE的容斥（P++注意）：

 #include <bits/stdc++.h>
 using namespace std;
 #define begin {
 #define end }
 #define while while(
 #define if if(
 #define do )
 #define then )
 #define for for(
 #define fillchar(a,b,c) memset(a,c,b)
 #define writeln printf("\n")
 #define write printf
 #define readln readl()
 #define inc(a) a++
 #define dec(a) a--
 #define exit(a) return a
 #define mod %
 #define div /
 #define shl <<
 #define shr >>
 #define extended long double
 #define longint int
 #define integer short
 #define int64 long long
 template<typename T> inline void read(T& a)
 begin
   T x=,f=; char ch=getchar();
   while(ch<'')or(ch>'')do
   begin
     if ch=='-' then f=-; ch=getchar();
   end
   while(ch>='')and(ch<='')do
   begin
     x=x*+ch-''; ch=getchar();
   end
   a=x*f;
 end
 inline void readl()
 begin
   char ch; ch=getchar();
   while ch!='\n' do ch=getchar();
 end
 int64 i,t,ii,j,n,m,x,a[],b[],ans;
 int main()
 begin
   read(t);
   for ii=;ii<=t;ii++ do
   begin
     read(x); j=sqrt(x); n=; m=; ans=;
     for i=;i<=j;i++ do
     begin
       if x mod i== then
       begin
         inc(n); a[n]=i;
         if x div i>i then begin inc(m); a[-m]=x div i; end;
       end
     end
     for i=n+;i<=n+m;i++ do a[i]=a[-(m-(i-n)+)];
     n=n+m;
     for i=;i<=n;i++ do b[a[i]]=;
     for i=n;i>=;i-- do
     begin
       b[a[i]]=b[a[i]]+(+x div a[i])*(x div a[i])div ;
       ans=ans+b[a[i]]*x;
       j=;
       while a[j]*a[j]<=a[i] do
       begin
         if j>n then break;
         if a[i] mod a[j]== then
         begin
           b[a[j]]=b[a[j]]-(a[i] div a[j])*b[a[i]];
           if(a[j]*a[j]<a[i])and(a[j]>)then
           b[a[i] div a[j]]=b[a[i] div a[j]]-a[j]*b[a[i]];
         end
         inc(j);
       end
     end
     write("%lld",ans); writeln;
   end
 end

标程（P++注意）：

 #include <bits/stdc++.h>
 using namespace std;
 #define begin {
 #define end }
 #define while while(
 #define if if(
 #define do )
 #define then )
 #define for for(
 #define fillchar(a,b,c) memset(a,c,b)
 #define writeln printf("\n")
 #define write printf
 #define readln readl()
 #define inc(a) a++
 #define dec(a) a--
 #define exit(a) return a
 #define mod %
 #define div /
 #define shl <<
 #define shr >>
 #define extended long double
 #define longint int
 #define integer short
 #define int64 long long
 template<typename T> inline void read(T& a)
 begin
   T x=,f=; char ch=getchar();
   while(ch<'')or(ch>'')do
   begin
     if ch=='-' then f=-; ch=getchar();
   end
   while(ch>='')and(ch<='')do
   begin
     x=x*+ch-''; ch=getchar();
   end
   a=x*f;
 end
 inline void readl()
 begin
   char ch; ch=getchar();
   while ch!='\n' do ch=getchar();
 end
 longint p[],vis[],ph[],pcnt=,T,n;
 void init_p()
 begin
   ph[]=; ph[]=;
   int64 temp;
   for int i=;i<;i++ do
   begin
     if not vis[i] then
     begin
       p[pcnt]=i; ph[i]=i-; inc(pcnt);
     end
     for int j=;j<pcnt&&(temp=(int64)p[j]*i)<;j++ do
     begin
       vis[temp]=;
       if i mod p[j]== then begin ph[temp]=ph[i]*p[j]; break; end
       else ph[temp]=ph[i]*(p[j]-);
     end
   end
 end
 int64 solve(int n)
 begin
   int64 ans=0ll;
   longint half=(int)(sqrt(n)+0.01);
   if half*half==n then begin ans+=1ll*ph[half]*half/; dec(half); end
   inc(ans); ans+=1ll*ph[n]*n/;
   for int i=;i<=half;i++ do
   if n mod i== then
   begin
     ans+=1ll*ph[i]*i/;
     ans+=1ll*ph[n/i]*n/i/;
   end
   exit(ans*n);
 }
 int main()
 begin
   read(T); init_p();
   for int i=;i<=T;i++ do
   begin read(n); write("%lld",solve(n)); writeln; end
   return ;
 end