cychester

BZOJ4407 于神之怒加强版 - 莫比乌斯反演

题解

非常裸的莫比乌斯反演。

但是反演完还需要快速计算一个积性函数(我直接用$nlogn$卷积被TLE了

推荐一个博客

我也不想再写一遍了

代码

 

 1 #include<cstring>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #define ll long long
 5 #define rd read()
 6 using namespace std;
 7  
 8 const ll mod = 1e9 + 7;
 9  
10 const int N = 5e6;
11  
12 int T, n, m, k;
13 int mu[N], pri[N], tot, vis[N], low[N];
14 ll sum[N];
15  
16 int read() {
17     int X = 0, p = 1; char c = getchar();
18     for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1;
19     for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0';
20     return X * p;
21 }
22  
23 ll fpow(ll a, ll b) {
24     ll re = 1;
25     for(; b; b >>= 1, a = a * a % mod) if(b & 1) re = re * a % mod;
26     return re;
27 }
28  
29 void init() {
30     mu[1] = 1;
31     sum[1] = 1;
32     for(int i = 2; i < N; ++i) {
33         if(!vis[i]) mu[i] = -1, pri[++tot] = i, sum[i] = fpow(i, k) - 1, low[i] = i;
34         for(int j = 1; j <= tot && i * pri[j] < N; ++j) {
35             vis[i * pri[j]] = 1;
36             if(i % pri[j] == 0) {
37                 low[i * pri[j]] = low[i] * pri[j];
38                 mu[i * pri[j]] = 0;
39                 if(low[i] == i) sum[i * pri[j]] = sum[i] * fpow(pri[j], k) % mod;
40                 else sum[i * pri[j]] =  sum[i / low[i]] * sum[pri[j] * low[i]] % mod;
41                 break;
42             }
43             low[i * pri[j]] = pri[j];
44             mu[i * pri[j]] = -mu[i];
45             sum[i * pri[j]] = sum[i] * sum[pri[j]] % mod;
46         }
47     }
48     //for(ll i = 1; i < N; ++i) {
49     //  ll tmp = fpow(i, k);
50     //  for(ll j = 1; j < N && i * j < N; ++j) sum[i * j] = (sum[i * j] + tmp * mu[j]) % mod;
51     //}
52     for(int i = 1; i < N; ++i) sum[i] = (sum[i - 1] + sum[i]) % mod;
53 }
54  
55 ll cal() {
56     ll ans = 0;
57     for(int i = 1, j = 1; i <= min(n, m); i = j + 1) {
58         j = n / (n / i);
59         j = min(j, m / (m / i));
60         ans += (sum[j] - sum[i - 1]) % mod * (n / i) % mod * (m / i) % mod;
61         ans %= mod;
62     }
63     ans = (ans % mod + mod) % mod;
64     return ans;
65 }
66  
67 int main()
68 {
69     T = rd; k = rd;
70     init();
71     for(; T; T--) {
72         n = rd; m = rd;
73         printf("%lld\n", cal());
74     }
75 }
View Code

 

posted on 2018-08-24 08:42  cychester  阅读(123)  评论(0编辑  收藏  举报

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