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：

 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<vector>
 #include<map>
 #include<set>
 #include<queue>
 #include<string>
 #define inf 1000000000
 #define maxn 1000000+5
 #define maxm 100000+5
 #define eps 1e-10
 #define ll long long
 #define pa pair<int,int>
 #define for0(i,n) for(int i=0;i<=(n);i++)
 #define for1(i,n) for(int i=1;i<=(n);i++)
 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
 #define mod 1000000007
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}
     return x*f;
 }
 int tot,p[maxn];
 ll fai[maxn];
 bool v[maxn];
 void get()
 {
     fai[]=;
     for2(i,,)
     {
         if(!v[i])p[++tot]=i,fai[i]=i-;
         for1(j,tot)
         {
             int k=i*p[j];
             if(k>)break;
             v[k]=;
             if(i%p[j])fai[k]=fai[i]*(p[j]-);
             else {fai[k]=fai[i]*p[j];break;}
         }
     }
     for2(i,,)(fai[i]*=(ll)i)>>=;
 }
 int main()
 {
     freopen("input.txt","r",stdin);
     freopen("output.txt","w",stdout);
     get();
     int T=read();
     while(T--)
     {
         int n=read(),m=sqrt(n);ll ans=;
         for1(i,m)if(n%i==)ans+=fai[n/i]+fai[i];
         if(m*m==n)ans-=fai[m];
         printf("%lld\n",ans*(ll)n);
     }
     return ;
 }

UPD：其实我们可以预处理出答案，用普通的筛法。

代码：

 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<vector>
 #include<map>
 #include<set>
 #include<queue>
 #include<string>
 #define inf 1000000000
 #define maxn 1000000+5
 #define maxm 1000000
 #define eps 1e-10
 #define ll long long
 #define pa pair<int,int>
 #define for0(i,n) for(int i=0;i<=(n);i++)
 #define for1(i,n) for(int i=1;i<=(n);i++)
 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
 #define mod 1000000007
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}
     return x*f;
 }
 int tot,p[maxn];
 ll fai[maxn],ans[maxn];
 bool v[maxn];
 void get()
 {
     fai[]=;
     for2(i,,maxm)
     {
         if(!v[i])p[++tot]=i,fai[i]=i-;
         for1(j,tot)
         {
             int k=i*p[j];
             if(k>maxm)break;
             v[k]=;
             if(i%p[j])fai[k]=fai[i]*(p[j]-);
             else {fai[k]=fai[i]*p[j];break;}
         }
     }
     for2(i,,maxm)(fai[i]*=(ll)i)>>=;
     for1(i,maxm)
      for(int j=i;j<=maxm;j+=i)
       ans[j]+=fai[i];
     for1(i,maxm)ans[i]*=(ll)i;
 }
 int main()
 {
     freopen("input.txt","r",stdin);
     freopen("output.txt","w",stdout);
     get();
     int T=read();
     while(T--)printf("%lld\n",ans[read()]);
     return ;
 }

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