bzoj 2226 LCMSum 欧拉函数

2226: [Spoj 5971] LCMSum

Time Limit: 20 Sec  Memory Limit: 259 MB
Submit: 1123  Solved: 492
[Submit][Status][Discuss]

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

思路：时间复杂度T*sqrt（n）；

　　　小于n并且与n互质的数为phi（k）*k/2；

　　　1的情况是特例；全部long long会超时。。。卡死我了

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define inf 999999999
#define esp 0.00000000001
const int MAXN=1000010;
int prime[MAXN],cnt;
bool vis[MAXN];
int phi[MAXN];
void Prime(int n)
{
    phi[1]=1;
    for(int i=2;i<n;i++)
    {
        if(!vis[i])
        {
            prime[cnt++]=i;
            phi[i]=i-1;
        }
        for(int j=0;j<cnt&&i*prime[j]<n;j++)
        {
            int k=i*prime[j];
            vis[k]=1;
            if(i%prime[j]==0)
            {
                phi[k]=phi[i]*prime[j];
                break;
            }
            else
            phi[k]=phi[i]*(prime[j]-1);
        }
    }
}
ll phii(ll n)
{
    return (ll)phi[n]*n/2;
}
int main()
{
    int x,y,z,i,t;
    Prime(1000001);
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&x);
        ll ans=0;
        ll gg=sqrt(x);
        for(i=1;i<=gg;i++)
        {
            if(x%i==0)
            ans+=phii(x/i),ans+=phii(i);
        }
        if(gg*gg==x)
        ans-=phii(gg);
        printf("%lld\n",(ans+1)*x);
    }
    return 0;
}