bzoj 4916 神犇和蒟蒻 杜教筛

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 }

标签：phi,4916,mo,long,杜教,int,MAXN,include,神犇
来源： https://www.cnblogs.com/iat14/p/11407908.html