bzoj 2226: [Spoj 5971] LCMSum 数论

2226: [Spoj 5971] LCMSum

Time Limit: 20 Sec  Memory Limit: 259 MB
Submit: 578  Solved: 259
[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

　　这道题将lcm转化为gcd并按照相同gcd分为一组的思路进行，巧妙地将题目转化为求小于等于n且与n互质数的和，而这个值时n*phi(n)/2

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAXN 1000010
typedef long long qword;
//segma(i*n/gcd(i,n))
//=n*segma(h(n/k)/k)
//h(a)表示与a互质数的和
bool pflag[MAXN];
int prime[MAXN],topp=-1;
int phi[MAXN];
void init()
{
        int i,j;
        int x,y;
        for (i=2;i<MAXN;i++)
        {
                if (!pflag[i])
                {
                        prime[++topp]=i;
                        phi[i]=i-1;
                }
                for (j=0;j<=topp && i*prime[j]<MAXN ;j++)
                {
                        pflag[i*prime[j]]=true;
                        phi[i*prime[j]]=phi[i]*phi[prime[j]];
                        if (i%prime[j]==0)
                        {
                                x=i;y=prime[j];
                                while (x%prime[j]==0)
                                {
                                        x/=prime[j];
                                        y*=prime[j];
                                }
                                if (x==1)
                                {
                                        phi[i*prime[j]]=i*(prime[j]-1);
                                }else
                                {
                                        phi[i*prime[j]]=phi[x]*phi[y];
                                }
                                continue;
                        }
                }
        }
}
qword h(int x)
{
        if (x==1)return 1;
        return (qword)x*phi[x]/2;
}
int main()
{
        freopen("input.txt","r",stdin);
        //freopen("output.txt","w",stdout);
        int nn;
        int n,i;
        scanf("%d",&nn);
        init();
        while (nn--)
        {
                scanf("%d",&n);
                qword ans=0;
                for (i=1;i*i<n;i++)
                {
                        if (n%i!=0)continue;
                        ans+=(qword)n*h(n/i);
                        ans+=(qword)n*h(i);
                }
                if (i*i==n)
                        ans+=(qword)n*h(i);
                printf("%lld
",ans);
        }
}