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lcm，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lcm[i]，并把更小的lcm[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lcm后，我们需要知道1~n div lcm-1中所有与n div lcm互质的数的和。
设m=n div lcm。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）：
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define begin {
4 #define end }
5 #define while while(
6 #define if if(
7 #define do )
8 #define then )
9 #define for for(
10 #define fillchar(a,b,c) memset(a,c,b)
11 #define writeln printf("\n")
12 #define write printf
13 #define readln readl()
14 #define inc(a) a++
15 #define dec(a) a--
16 #define exit(a) return a
17 #define mod %
18 #define div /
19 #define shl <<
20 #define shr >>
21 #define extended long double
22 #define longint int
23 #define integer short
24 #define int64 long long
25 template<typename T> inline void read(T& a)
26 begin
27 T x=0,f=1; char ch=getchar();
28 while(ch<‘0‘)or(ch>‘9‘)do
29 begin
30 if ch==‘-‘ then f=-1; ch=getchar();
31 end
32 while(ch>=‘0‘)and(ch<=‘9‘)do
33 begin
34 x=x*10+ch-‘0‘; ch=getchar();
35 end
36 a=x*f;
37 end
38 inline void readl()
39 begin
40 char ch; ch=getchar();
41 while ch!=‘\n‘ do ch=getchar();
42 end
43 int64 i,t,ii,j,n,m,x,a[3002],b[1000002],ans;
44 int main()
45 begin
46 read(t);
47 for ii=1;ii<=t;ii++ do
48 begin
49 read(x); j=sqrt(x); n=0; m=0; ans=0;
50 for i=1;i<=j;i++ do
51 begin
52 if x mod i==0 then
53 begin
54 inc(n); a[n]=i;
55 if x div i>i then begin inc(m); a[2002-m]=x div i; end;
56 end
57 end
58 for i=n+1;i<=n+m;i++ do a[i]=a[2002-(m-(i-n)+1)];
59 n=n+m;
60 for i=1;i<=n;i++ do b[a[i]]=0;
61 for i=n;i>=1;i-- do
62 begin
63 b[a[i]]=b[a[i]]+(1+x div a[i])*(x div a[i])div 2;
64 ans=ans+b[a[i]]*x;
65 j=1;
66 while a[j]*a[j]<=a[i] do
67 begin
68 if j>n then break;
69 if a[i] mod a[j]==0 then
70 begin
71 b[a[j]]=b[a[j]]-(a[i] div a[j])*b[a[i]];
72 if(a[j]*a[j]<a[i])and(a[j]>1)then
73 b[a[i] div a[j]]=b[a[i] div a[j]]-a[j]*b[a[i]];
74 end
75 inc(j);
76 end
77 end
78 write("%lld",ans); writeln;
79 end
80 end
标程（P++注意）：
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define begin {
4 #define end }
5 #define while while(
6 #define if if(
7 #define do )
8 #define then )
9 #define for for(
10 #define fillchar(a,b,c) memset(a,c,b)
11 #define writeln printf("\n")
12 #define write printf
13 #define readln readl()
14 #define inc(a) a++
15 #define dec(a) a--
16 #define exit(a) return a
17 #define mod %
18 #define div /
19 #define shl <<
20 #define shr >>
21 #define extended long double
22 #define longint int
23 #define integer short
24 #define int64 long long
25 template<typename T> inline void read(T& a)
26 begin
27 T x=0,f=1; char ch=getchar();
28 while(ch<‘0‘)or(ch>‘9‘)do
29 begin
30 if ch==‘-‘ then f=-1; ch=getchar();
31 end
32 while(ch>=‘0‘)and(ch<=‘9‘)do
33 begin
34 x=x*10+ch-‘0‘; ch=getchar();
35 end
36 a=x*f;
37 end
38 inline void readl()
39 begin
40 char ch; ch=getchar();
41 while ch!=‘\n‘ do ch=getchar();
42 end
43 longint p[1000100],vis[1000100],ph[1000100],pcnt=0,T,n;
44 void init_p()
45 begin
46 ph[1]=1; ph[2]=1;
47 int64 temp;
48 for int i=2;i<1000100;i++ do
49 begin
50 if not vis[i] then
51 begin
52 p[pcnt]=i; ph[i]=i-1; inc(pcnt);
53 end
54 for int j=0;j<pcnt&&(temp=(int64)p[j]*i)<1000100;j++ do
55 begin
56 vis[temp]=1;
57 if i mod p[j]==0 then begin ph[temp]=ph[i]*p[j]; break; end
58 else ph[temp]=ph[i]*(p[j]-1);
59 end
60 end
61 end
62 int64 solve(int n)
63 begin
64 int64 ans=0ll;
65 longint half=(int)(sqrt(n)+0.01);
66 if half*half==n then begin ans+=1ll*ph[half]*half/2; dec(half); end
67 inc(ans); ans+=1ll*ph[n]*n/2;
68 for int i=2;i<=half;i++ do
69 if n mod i==0 then
70 begin
71 ans+=1ll*ph[i]*i/2;
72 ans+=1ll*ph[n/i]*n/i/2;
73 end
74 exit(ans*n);
75 }
76 int main()
77 begin
78 read(T); init_p();
79 for int i=1;i<=T;i++ do
80 begin read(n); write("%lld",solve(n)); writeln; end
81 return 0;
82 end
时间： 2024-11-10 01:22:01