BZOJ2226: [Spoj 5971] LCMSum

题解：

考虑枚举gcd，然后问题转化为求<=n且与n互质的数的和。

这是有公式的f[i]=phi[i]*i/2

然后卡一卡时就可以过了。

代码：

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<iostream>
 7 #include<vector>
 8 #include<map>
 9 #include<set>
10 #include<queue>
11 #include<string>
12 #define inf 1000000000
13 #define maxn 1000000+5
14 #define maxm 100000+5
15 #define eps 1e-10
16 #define ll long long
17 #define pa pair<int,int>
18 #define for0(i,n) for(int i=0;i<=(n);i++)
19 #define for1(i,n) for(int i=1;i<=(n);i++)
20 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
21 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
22 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
23 #define mod 1000000007
24 using namespace std;
25 inline int read()
26 {
27     int x=0,f=1;char ch=getchar();
28     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
29     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
30     return x*f;
31 }
32 int tot,p[maxn];
33 ll fai[maxn];
34 bool v[maxn];
35 void get()
36 {
37     fai[1]=1;
38     for2(i,2,1000000)
39     {
40         if(!v[i])p[++tot]=i,fai[i]=i-1;
41         for1(j,tot)
42         {
43             int k=i*p[j];
44             if(k>1000000)break;
45             v[k]=1;
46             if(i%p[j])fai[k]=fai[i]*(p[j]-1);
47             else {fai[k]=fai[i]*p[j];break;}
48         }
49     }
50     for2(i,3,1000000)(fai[i]*=(ll)i)>>=1;
51 }
52 int main()
53 {
54     freopen("input.txt","r",stdin);
55     freopen("output.txt","w",stdout);
56     get();
57     int T=read();
58     while(T--)
59     {
60         int n=read(),m=sqrt(n);ll ans=0;
61         for1(i,m)if(n%i==0)ans+=fai[n/i]+fai[i];
62         if(m*m==n)ans-=fai[m];
63         printf("%lld
",ans*(ll)n);
64     }
65     return 0;
66 }

View Code

UPD：其实我们可以预处理出答案，用普通的筛法。

代码：

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<iostream>
 7 #include<vector>
 8 #include<map>
 9 #include<set>
10 #include<queue>
11 #include<string>
12 #define inf 1000000000
13 #define maxn 1000000+5
14 #define maxm 1000000
15 #define eps 1e-10
16 #define ll long long
17 #define pa pair<int,int>
18 #define for0(i,n) for(int i=0;i<=(n);i++)
19 #define for1(i,n) for(int i=1;i<=(n);i++)
20 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
21 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
22 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
23 #define mod 1000000007
24 using namespace std;
25 inline int read()
26 {
27     int x=0,f=1;char ch=getchar();
28     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
29     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
30     return x*f;
31 }
32 int tot,p[maxn];
33 ll fai[maxn],ans[maxn];
34 bool v[maxn];
35 void get()
36 {
37     fai[1]=1;
38     for2(i,2,maxm)
39     {
40         if(!v[i])p[++tot]=i,fai[i]=i-1;
41         for1(j,tot)
42         {
43             int k=i*p[j];
44             if(k>maxm)break;
45             v[k]=1;
46             if(i%p[j])fai[k]=fai[i]*(p[j]-1);
47             else {fai[k]=fai[i]*p[j];break;}
48         }
49     }
50     for2(i,3,maxm)(fai[i]*=(ll)i)>>=1;
51     for1(i,maxm)
52      for(int j=i;j<=maxm;j+=i)
53       ans[j]+=fai[i];
54     for1(i,maxm)ans[i]*=(ll)i;
55 }
56 int main()
57 {
58     freopen("input.txt","r",stdin);
59     freopen("output.txt","w",stdout);
60     get();
61     int T=read();
62     while(T--)printf("%lld
",ans[read()]);
63     return 0;
64 }

View Code

2226: [Spoj 5971] LCMSum

Time Limit: 20 Sec Memory Limit: 259 MB
Submit: 659 Solved: 292
[Submit][Status]

Description

Given n, calculate the sum LCM(1,n) + LCM(2,n) + .. + LCM(n,n), where LCM(i,n) denotes the Least Common Multiple of the integers i and n.

Input

The first line contains T the number of test cases. Each of the next T lines contain an integer n.

Output

Output T lines, one for each test case, containing the required sum.

Sample Input

3
1
2
5

Sample Output

1
4
55

HINT

Constraints

1 <= T <= 300000
1 <= n <= 1000000