BZOJ4916 神犇和蒟蒻

题意：求miu[i*i]和phi[i*i]的前缀和。n <= 1e9

解：杜教筛。

第一问是搞笑的，答案是1。

第二问，发现phi[i*i] = phi[i] * i

这里杜教筛用的g(n) = n，h(n) = n²。

然后套路一波就没了。注意预处理的时候也要开long long

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 typedef long long LL;
 5 const int N = 1100010, T = 1100008;
 6 const LL MO = 1e9 + 7, inv6 = 166666668;
 7 
 8 int p[N], top, phi[N];
 9 LL F[N], F2[N], n, inv2;
10 bool vis[N];
11 
12 inline LL H(LL x) {
13     x %= MO;
14     return (x << 1 | 1) % MO * x % MO * (x + 1) % MO * inv6 % MO;
15 }
16 
17 inline void getp(int n) {
18     phi[1] = 1;
19     for(int i = 2; i <= n; i++) {
20         if(!vis[i]) {
21             p[++top] = i;
22             phi[i] = i - 1;
23         }
24         for(int j = 1; j <= top && i * p[j] <= n; j++) {
25             vis[i * p[j]] = 1;
26             if(i % p[j] == 0) {
27                 phi[i * p[j]] = phi[i] * p[j];
28                 break;
29             }
30             phi[i * p[j]] = phi[i] * (p[j] - 1);
31         }
32     }
33     for(int i = 1; i <= n; i++) {
34         F[i] = (F[i - 1] + 1ll * i * phi[i] % MO) % MO;
35     }
36     return;
37 }
38 
39 LL solve(LL x) {
40     //printf("solve %lld 
", x);
41     if(x <= 0) return 0;
42     if(x <= T) return F[x];
43     if(F2[n / x] != -1) return F2[n / x];
44     //printf("solve %lld 
", x);
45     LL ans = H(x);
46     for(LL i = 2, j; i <= x; i = j + 1) {
47         j = x / (x / i);
48         ans -= (i + j) % MO * ((j - i + 1) % MO) % MO * inv2 % MO * solve(x / i) % MO;
49         ans = (ans % MO + MO) % MO;
50     }
51     return F2[n / x] = ans;
52 }
53 
54 int main() {
55     inv2 = (MO + 1) / 2;
56     memset(F2, -1, sizeof(F2));
57     getp(T);
58     scanf("%lld", &n);
59     printf("1
%lld
", solve(n));
60     return 0;
61 }

AC代码