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("
")
#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=0,f=1; char ch=getchar();
  while(ch<'0')or(ch>'9')do
  begin
    if ch=='-' then f=-1; ch=getchar();
  end
  while(ch>='0')and(ch<='9')do
  begin
    x=x*10+ch-'0'; ch=getchar();
  end
  a=x*f;
end
inline void readl()
begin
  char ch; ch=getchar();
  while ch!='
' do ch=getchar();
end
int64 i,t,ii,j,n,m,x,a[3002],b[1000002],ans;
int main()
begin
  read(t);
  for ii=1;ii<=t;ii++ do
  begin
    read(x); j=sqrt(x); n=0; m=0; ans=0;
    for i=1;i<=j;i++ do
    begin
      if x mod i==0 then
      begin
        inc(n); a[n]=i;
        if x div i>i then begin inc(m); a[2002-m]=x div i; end;
      end
    end
    for i=n+1;i<=n+m;i++ do a[i]=a[2002-(m-(i-n)+1)];
    n=n+m;
    for i=1;i<=n;i++ do b[a[i]]=0;
    for i=n;i>=1;i-- do
    begin
      b[a[i]]=b[a[i]]+(1+x div a[i])*(x div a[i])div 2;
      ans=ans+b[a[i]]*x;
      j=1;
      while a[j]*a[j]<=a[i] do
      begin
        if j>n then break;
        if a[i] mod a[j]==0 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]>1)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("
")
#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=0,f=1; char ch=getchar();
  while(ch<'0')or(ch>'9')do
  begin
    if ch=='-' then f=-1; ch=getchar();
  end
  while(ch>='0')and(ch<='9')do
  begin
    x=x*10+ch-'0'; ch=getchar();
  end
  a=x*f;
end
inline void readl()
begin
  char ch; ch=getchar();
  while ch!='
' do ch=getchar();
end
longint p[1000100],vis[1000100],ph[1000100],pcnt=0,T,n;
void init_p() 
begin
  ph[1]=1; ph[2]=1;
  int64 temp;
  for int i=2;i<1000100;i++ do
  begin
    if not vis[i] then
    begin
      p[pcnt]=i; ph[i]=i-1; inc(pcnt);
    end 
    for int j=0;j<pcnt&&(temp=(int64)p[j]*i)<1000100;j++ do
    begin
      vis[temp]=1;
      if i mod p[j]==0 then begin ph[temp]=ph[i]*p[j]; break; end
      else ph[temp]=ph[i]*(p[j]-1);
    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/2; dec(half); end
  inc(ans); ans+=1ll*ph[n]*n/2;
  for int i=2;i<=half;i++ do
  if n mod i==0 then
  begin
    ans+=1ll*ph[i]*i/2;
    ans+=1ll*ph[n/i]*n/i/2;
  end
  exit(ans*n);
}
int main() 
begin
  read(T); init_p();
  for int i=1;i<=T;i++ do
  begin read(n); write("%lld",solve(n)); writeln; end
  return 0;
end