51nod 1584加权约数和

学到了好多东西啊这题。。。

https://blog.csdn.net/sdfzyhx/article/details/72968468

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=1e6+10,mod=1e9+7;
 4 typedef long long ll;
 5 ll miu[N],v[N],p[N],cnt,sum1[N],sum2[N],mi1[N],mi2[N],h[N],g[N],f[N],T,n;
 6 ll inc(ll a,ll b)
 7 {
 8     a=a+b;
 9     if(a>mod)a%=mod;
10     return a;
11 }
12 ll dec(ll a,ll b)
13 {
14     a=a-b;
15     if(a<0)a+=mod;
16     return a%mod;
17 }
18 ll qmod(ll x,ll y)
19 {
20     int ans=1;
21     while(y)
22     {
23         if(y&1)ans=1ll*ans*x%mod;
24         x=1ll*x*x%mod;y>>=1;
25     }
26     return ans;
27 }
28 void init()
29 {
30     miu[1]=sum1[1]=sum2[1]=mi1[1]=mi2[1]=1;
31     for(int i=2;i<=1e6;++i)
32     {
33         if(!v[i])
34         {
35             p[++cnt]=i;
36             miu[i]=-1;
37             sum1[i]=i+1;
38             sum2[i]=inc(1ll*i*i%mod+i,1);
39             mi1[i]=mi2[i]=i;
40         }
41         for(int j=1;j<=cnt&&i*p[j]<=1e6;++j)
42         {
43             int x=i*p[j];v[x]=1;
44             if(i%p[j])
45             {
46                 mi1[x]=mi2[x]=p[j];
47                 miu[x]=-miu[i];
48                 sum1[x]=1ll*sum1[i]*sum1[p[j]]%mod;
49                 sum2[x]=1ll*sum2[i]*sum2[p[j]]%mod;
50             }
51             else
52             {
53                 mi1[x]=p[j];
54                 mi2[x]=1ll*mi2[i]*p[j]%mod;
55                 sum1[x]=inc(sum1[i],1ll*p[j]*mi2[i]%mod*sum1[i/mi2[i]]%mod);//求约数和 
56                 sum2[x]=inc(sum2[i],1ll*inc(1ll*mi2[i]*mi2[i]%mod*mi1[i]%mod,1ll*mi2[i]*mi2[i]%mod*mi1[i]%mod*mi1[i]%mod)*sum2[i/mi2[i]]%mod);//求平方约数和 
57                 break;
58             }
59         }
60     }
61     for(int i=1;i<=1e6;++i)
62     {
63         h[i]=inc(h[i-1],sum1[i]);
64         g[i]=1ll*i*sum1[i]%mod*h[i]%mod;
65         sum2[i]=1ll*sum2[i]*i%mod;
66         sum2[i]=inc(sum2[i],sum2[i-1]);
67     }    
68     for(int i=1;i<=1e6;++i)
69     {
70 
71         for(int j=i,k=1;j<=1e6;j+=i,k++)
72         f[j]=inc(f[j],1ll*miu[i]%mod*i%mod*i%mod*g[k]%mod);
73         f[i]=inc(f[i],f[i-1]);
74     }
75     return;
76 }
77 int main()
78 {
79     init();scanf("%d",&T);
80     for(int i=1;i<=T;++i)
81     {
82         scanf("%d",&n);
83         printf("Case #%d: %lld\n",i,(dec(2ll*f[n]%mod,sum2[n])+mod)%mod);
84     }
85     return 0;
86 }

 

posted @ 2018-03-23 17:40  大奕哥&VANE  阅读(330)  评论(0)  编辑  收藏  举报