P1390 公约数的和

首杀莫反。

$\text{Problem}$

推式子。

其中:

AC Code
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
#include<bits/stdc++.h>
#define gh() getchar()
#define re register
typedef long long ll;
template<class T>
inline void read(T &x)
{
x=0;
char ch=gh(),t=0;
while(ch<'0'||ch>'9') t|=ch=='-',ch=gh();
while(ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+(ch^48),ch=gh();
if(t) x=-x;
}
template<class T,class ...T1>
inline void read(T &x,T1 &...x1)
{
read(x),read(x1...);
}
template<class T>
inline void write(T x)
{
if(x<0) putchar('-'),x=-x;
if(x>9) write(x/10);
putchar(x%10+'0');
}
template<>
inline void write(char c)
{
putchar(c);
}
template<>
inline void write(char *s)
{
while(*s) putchar(*s++);
}
template<class T,class ...T1>
inline void write(T x,T1 ...x1)
{
write(x),write(x1...);
}
template<class T>
inline bool checkMax(T &x,T &y)
{
return x<y?x=y,1:0;
}
template<class T>
inline bool checkMin(T &x,T &y)
{
return x>y?x=y,1:0;
}
const int MAXN=2e6+10;
int N,Tot;
int phi[MAXN],pri[MAXN];
bool is[MAXN];
inline void sieve(int n)
{
phi[1]=1;
for(int i=2;i<=n;++i)
{
if(!is[i]) pri[++Tot]=i,phi[i]=i-1;
for(int j=1;j<=Tot&&i*pri[j]<=n;++j)
{
is[i*pri[j]]=1;
if(i%pri[j]==0)
{
phi[i*pri[j]]=phi[i]*pri[j];
break;
}
phi[i*pri[j]]=phi[i]*(pri[j]-1);
}
}
}
inline void solve(int n)
{
ll res=0;
for(int i=1;i<=n;++i) res+=1ll*phi[i]*(n/i)*(n/i);
res-=1ll*(n+1)*n/2;
res>>=1;
write(res);
}
int main()
{
// freopen("gcd.in","r",stdin);
// freopen("gcd.out","w",stdout);
read(N);
sieve(N+9);
solve(N);
return 0;
}
/*
10
*/