Description

給出一個數字N

Input

Output

Sample Input

Sample Output

HINT

Source

By FancyCoder

 1 #include<cstring>
 2 #include<cmath>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cstdio>
 6 
 7 #define ll long long
 8 #define N 10000007
 9 using namespace std;
10 inline int read()
11 {
12     int x=0,f=1;char ch=getchar();
13     while
 
(ch<‘0‘||ch>‘9‘){if (ch==‘-‘)f=-1;ch=getchar();}
14     while(ch>=‘0‘&&ch<=‘9‘){x=(x<<1)+(x<<3)+ch-‘0‘;ch=getchar();}
15     return x*f;
16 }
17 
18 int n;
19 int cnt,pri[N],phi[N];
20 ll sum[N];
21 bool flag[N];
22 
23 void init_phi()
24 {
25     phi[1]=1;
26
 
     for (int i=2;i<=10000000;i++)
27     {
28         if (!flag[i])pri[++cnt]=i,phi[i]=i-1;
29         for (int j=1;j<=cnt&&pri[j]*i<=10000000;j++)
30         {
31             flag[pri[j]*i]=1;
32             if (i%pri[j]==0)
33             {
34                 phi[pri[j]*i]=pri[j]*phi[i];
35                 break;
36             }
37             else phi[pri[j]*i]=phi[i]*phi[pri[j]];
38         }
39     }
40     for (int i=1;i<=10000000;i++)sum[i]=sum[i-1]+phi[i];
41 }
42 void solve(ll n)
43 {
44     ll last,ans=0;
45     for (int i=1;i<=n;i=last+1)
46     {
47         last=n/(n/i);//琛ㄧずn涓埌杈懼摢閲屼負姝㈤兘鏄痭/i鐨?
48         ans+=(sum[last]-sum[i-1])*((sum[n/i]<<1)-1);
49     }
50     printf("%lld\n",ans);
51 }
52 int main()
53 {
54     init_phi();
55     int T=read();
56     while(T--)
57     {
58         n=read();
59         solve(n);
60     }
61 }

bzoj 4804 歐拉心算 歐拉函數,莫比烏斯