第十九周 1.3-1.9

1.3

补个BC。

HDU 5608 function

题解这么说。复杂度不会。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <map>
 4 using namespace std;
 5 typedef long long LL;
 6 const LL mod = 1e9 + 7;
 7 const LL maxn = 1e6 + 10;
 8 map<LL,LL> M;
 9 LL t[maxn];
10
11 LL S1(LL n)
12 {
13     LL ret = n * ( n + 1 ) % mod * ( 2 * n % mod + 1 ) % mod * 166666668LL % mod;
14     ret = ( ret - 3 * n % mod * ( n + 1 ) % mod * 500000004LL % mod + mod ) % mod;
15     ret = ( ret + 2 * n ) % mod;
16     return ret;
17 }
18
19 LL S2(LL n)
20 {
21     if(n < maxn) return t[n];
22     if(M.find(n) != M.end()) return M[n];
23     LL ret = S1(n);
24     for(LL i = 2; i <= n; i++)
25     {
26         LL r = n / ( n / i );
27         ret = (ret - S2(n/i) * ( r - i + 1 ) % mod + mod) % mod;
28         i = r;
29     }
30     return M[n] = ret;
31 }
32
33 void pre()
34 {
35     for(LL i = 1; i < maxn; i++) t[i] = (i - 1) * (i - 2) % mod;
36     for(int i = 1; i < maxn; i++)
37         for(int j = 2; i * j < maxn; j++)
38             t[i*j] = (t[i*j] - t[i] + mod) % mod;
39     for(int i = 1; i < maxn; i++) t[i] = ( t[i] + t[i-1] ) % mod;
40     return;
41 }
42
43 int main(void)
44 {
45     pre();
46     int T;
47     scanf("%d", &T);
48     while(T--)
49     {
50         LL n;
51         scanf("%I64d", &n);
52         printf("%I64d\n", S2(n));
53     }
54     return 0;
55 }

Aguin