题目链接：点击这里
题目大意：
给定一个正整数
N
N
N ，对于所有的
n
∈
[
1
,
N
]
n\in[1,N]
n∈[1,N] 输出
∑
i
=
1
n
i
gcd
(
i
,
n
)
\sum_{i=1}^n\frac{i}{\gcd(i,n)}
∑i=1ngcd(i,n)i
题目分析：
∑
i
=
1
n
i
gcd
(
i
,
n
)
\sum_{i=1}^n\frac{i}{\gcd(i,n)}
i=1∑ngcd(i,n)i
=
∑
d
=
1
n
1
d
∑
i
=
1
n
i
[
gcd
(
i
,
n
)
=
d
]
=\sum_{d=1}^n\frac{1}{d}\sum_{i=1}^n i[\gcd(i,n)=d]
=d=1∑nd1i=1∑ni[gcd(i,n)=d]
=
∑
d
=
1
n
1
d
∑
i
=
1
⌊
n
d
⌋
i
d
[
gcd
(
i
,
n
/
d
)
=
1
]
=\sum_{d=1}^n\frac{1}{d}\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor} id[\gcd(i,n/d)=1]
=d=1∑nd1i=1∑⌊dn⌋id[gcd(i,n/d)=1]
=
∑
d
=
1
n
n
/
d
⋅
φ
(
n
/
d
)
+
[
n
/
d
=
1
]
2
=\sum_{d=1}^n\frac{n/d·\varphi(n/d)+[n/d=1]}{2}
=d=1∑n2n/d⋅φ(n/d)+[n/d=1]
∑
d
∣
n
d
⋅
φ
(
d
)
+
[
d
=
1
]
2
\sum_{d|n}\frac{d·\varphi(d)+[d=1]}{2}
d∣n∑2d⋅φ(d)+[d=1]
提前筛一下欧拉函数，然后根据上述公式调和级数枚举枚举因子计算贡献即可
具体细节见代码：
//#pragma GCC optimize(2)
//#pragma GCC optimize("Ofast","inline","-ffast-math")
//#pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<iomanip>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<queue>
#define ll long long
#define inf 0x3f3f3f3f
#define Inf 0x3f3f3f3f3f3f3f3f
#define int ll
#define endl '\n'
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
using namespace std;
int read()
{
int res = 0,flag = 1;
char ch = getchar();
while(ch<'0' || ch>'9')
{
if(ch == '-') flag = -1;
ch = getchar();
}
while(ch>='0' && ch<='9')
{
res = (res<<3)+(res<<1)+(ch^48);//res*10+ch-'0';
ch = getchar();
}
return res*flag;
}
const int maxn = 1e6+5;
const int mod = 1e9+7;
const double pi = acos(-1);
const double eps = 1e-8;
int n,cnt,pri[maxn],phi[maxn],ans[maxn];
bool vis[maxn];
void init(int n)
{
vis[1] = phi[1] = 1;
for(int i = 2;i <= n;i++)
{
if(!vis[i])
{
pri[++cnt] = i;
phi[i] = i-1;
}
for(int j = 1;j<=cnt && i*pri[j]<=n;j++)
{
vis[i*pri[j]] = true;
if(i%pri[j] == 0)
{
phi[i*pri[j]] = phi[i]*pri[j];
break;
}
phi[i*pri[j]] = phi[i]*(pri[j]-1);
}
}
}
signed main()
{
n = read();
init(n);
for(int i = 1;i <= n;i++)
for(int j = 1;j*i <= n;j++) ans[i*j] += phi[j]*j/2;
for(int i = 1;i <= n;i++) cout<<ans[i]+1<<endl;
return 0;
}