第一题结合莫比乌斯函数定义，值恒为1。
第二题，phi(i^2) = phi(i) * i，根据欧拉函数的定义式能推出来，每个质因子的指数都增加一倍，都提出来一份，就是原先的phi(i)*i。然后还是跟g(x)卷一下，杜教筛即可。
1 #include <cstdio>
2 #include <map>
3 #include <cmath>
4 using namespace std;
5 typedef long long ll;
6 const int MAXN = 1000100,mo = 1e9 + 7,inv2 = (mo + 1) / 2,inv6 = (mo + 1) / 6;
7
8 int n,maxn,phi[MAXN],sum[MAXN],pri[MAXN];
9 bool vis[MAXN];
10 map<int,int> f;
11 int solve(int n)
12 {
13 if (n <= maxn)
14 return sum[n];
15 if (f.count(n))
16 return f[n];
17 int ans = (ll)n * (n + 1) % mo * (n * 2 + 1) % mo * inv6 % mo;
18 for (int l = 2,r;l <= n;l = r + 1)
19 {
20 r = n / (n / l);
21 int tp = (ll)(l + r) * (r - l + 1) % mo * inv2 % mo;
22 ans -= (ll)tp * solve(n / l) % mo;
23 if (ans < 0)
24 ans += mo;
25 }
26 return f[n] = ans;
27 }
28 void init()
29 {
30 maxn = pow(n,2.0 / 3) + 0.5;
31 phi[1] = 1;
32 int tot = 0;
33 for (int i = 2;i <= maxn;i++)
34 {
35 if (vis[i] == false)
36 {
37 pri[++tot] = i;
38 phi[i] = i - 1;
39 }
40 for (int j = 1;j <= tot && i * pri[j] <= maxn;j++)
41 {
42 vis[i * pri[j]] = true;
43 if (i % pri[j] != 0)
44 phi[i * pri[j]] = phi[i] * (pri[j] - 1);
45 else
46 {
47 phi[i * pri[j]] = phi[i] * pri[j];
48 break;
49 }
50 }
51 }
52 for (int i = 1;i <= maxn;i++)
53 sum[i] = (sum[i - 1] + (ll)phi[i] * i % mo) % mo;
54 }
55
56 int main()
57 {
58 scanf("%d",&n);
59 init();
60 printf("1\n%d\n",solve(n));
61 return 0;
62 }