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