【SPOJ】 LCMSUM（数论函数+卷积）

【spoj 】LCMSUM

 

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.

Example

Sample Input:
3
1
2
5

Sample Output:
1
4
55

Constraints

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

 

#include<cstdio>
#include<iostream>
#define ll long long
using namespace std;
const int MAXN = 1000010;
int phi[MAXN], pri[MAXN];
ll ans[MAXN];
bool vis[MAXN];
void init()
{
    phi[1] = 1;
    int cnt = 0;
    for (int i = 2; i < MAXN; i++)
    {
        if (!vis[i])
            pri[++cnt] = i, phi[i] = i - 1,vis[i] = 1;
        for (int j = 1; j <= cnt && i*pri[j] < MAXN; j++)
        {
            vis[i*pri[j]] = 1;
            if (i%pri[j] == 0)
            {
                phi[i*pri[j]] = phi[i] * pri[j];
                break;
            }
            phi[i*pri[j]] = phi[i] * (pri[j] - 1);
        }
        
    }for (int i = 1; i < MAXN; i++)
            for (int j = 1; i*j < MAXN; j++)
                ans[i*j] += 1LL * j*phi[j]/2;
    for (int i = 1; i < MAXN; i++)
        ans[i] = i * ans[i] + i;
}
int main()
{
    int T;
    init();
    scanf("%d", &T);
    while (T--)
    {
        int n;
        scanf("%d", &n);
        printf("%lld\n", ans[n]);
    }
    return 0;
}