[欧拉函数]BZOJ 2226 LCMsum

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

分析

原式：$\sum _{i=1}^n lcm(i,n)$

$n\sum _{i=1}^n \frac{i}{gcd(i,n)}$

考虑直接枚举因子：

$n\sum _{d|n} \sum _{i=1}^{\frac{n}{d}} [gcd(\frac{n}{d},i)=1]i$

$n\sum _{d|n} \frac{d\varphi (d)}{2}$

线性筛预处理phi即可

 

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
const int N=1e6+1;
int T,n;
bool nonprime[N];
int prime[N],cnt,phi[N];
ll ans[N];

void Solve() {
    phi[1]=1;
    for (int i=1;i<N;i++) ans[i]=1;
    for (int i=2;i<N;i++) {
        if (!nonprime[i]) prime[++cnt]=i,phi[i]=i-1;
        for (int j=i;j<N;j+=i) ans[j]+=1ll*i*phi[i]>>1;
        for (int j=1;j<=cnt;j++) {
            if (1ll*i*prime[j]>=N) break;
            nonprime[i*prime[j]]=1;
            if (i%prime[j]==0) {
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
            phi[i*prime[j]]=phi[i]*phi[prime[j]];
        }
    }
}

int main() {
    Solve();
    for (scanf("%d",&T);T;T--) {
        scanf("%d",&n);
        printf("%lld\n",ans[n]*n);
    }
}

 

来源：https://www.cnblogs.com/mastervan/p/11367350.html