[BZOJ 4916]神犇和蒟蒻

Description

很久很久以前，有一只神犇叫yzy;

很久很久之后，有一只蒟蒻叫lty;

Input

请你读入一个整数N;1<=N<=1E9,A、B模1E9+7;

Output

请你输出一个整数 $A=sum_{i=1}^N{mu (i^2)}$ ;

请你输出一个整数 $B=sum_{i=1}^N{varphi (i^2)}$ ;

Sample Input

1

Sample Output

1
1

题解

首先注意到 $A$ 直接输出 $1$ 得满分。因为只有 $mu(1^2)=1$ 。至于为什么，想想莫比乌斯函数是什么。

对于第二问容易发现 $varphi(i^2)=icdotvarphi(i)$ 。至于为什么，想想你是怎么线性筛欧拉函数的。

对于函数 $f(i)=icdotvarphi(i)$ 容易发现，这也是个积性函数。我们可以将阈值内的线性筛筛出来。对于阈值外的，考虑杜教筛。

求 $S(n)=sumlimits_{i=1}^nf(i)$

上述式子 $$g(1)S(n)=sum_{i=1}^n(g*f)(i)-sum_{i=2}^ng(i)Sleft(leftlfloorfrac{n}{i} ight floor ight)$$

考虑到 $sumlimits_{dmid n}varphi(d)=n$ ，又由于 $(g*f)(n)=sumlimits_{dmid n}varphi(d)dcdot gleft(frac{n}{d} ight)$ 。我们考虑让 $g(n)=id(n)$ ，那么 $(id*f)(n)=sumlimits_{dmid n}ncdotvarphi(d)=n^2$ 。由于 $sumlimits_{i=1}^ni^2=frac{n(n+1)(2n+1)}{6}$ 。显然这个卷积的前缀为 $sumlimits_{i=1}^n(g*f)(i)=frac{n(n+1)(2n+1)}{6}$ 。

故对于 $f$ $$S(n)=frac{n(n+1)(2n+1)}{6}-sum_{i=2}^nicdot Sleft(leftlfloorfrac{n}{i} ight floor ight)$$

//It is made by Awson on 2018.1.24
#include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('
'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int MOD = 1e9+7;
const int N = 2333333;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void write(int x) {
    if (x > 9) write(x/10);
    putchar(x%10+48);
}

int n, inv2, inv6, f[N+5];
int prime[N+5], isprime[N+5], tot;
map<int, int>mp;

int quick_pow(int a, int b) {
    int ans = 1;
    while (b) {
    if (b&1) ans = 1ll*ans*a%MOD;
    a = 1ll*a*a%MOD, b >>= 1;
    }
    return ans;
}
void get_f() {
    memset(isprime, 1, sizeof(isprime)); isprime[1] = 0, f[1] = 1;
    for (int i = 2; i <= N; i++) {
    if (isprime[i]) prime[++tot] = i, f[i] = 1ll*i*(i-1)%MOD;
    for (int j = 1; j <= tot && i*prime[j] <= N; j++) {
        isprime[i*prime[j]] = 0;
        if (i%prime[j]) f[i*prime[j]] = 1ll*f[i]*(prime[j]-1)%MOD*prime[j]%MOD;
        else {f[i*prime[j]] = 1ll*f[i]*prime[j]%MOD*prime[j]%MOD; break; }
    }
    (f[i] += f[i-1]) %= MOD;
    }
}
int cal(int x) {
    if (x <= N) return f[x];
    if (mp.count(x)) return mp[x];
    int ans = 1ll*x*(x+1)%MOD*(2*x+1)%MOD*inv6%MOD;
    for (int i = 2, last; i <= x; i = last+1) {
    last = x/(x/i); (ans -= 1ll*(last+i)*(last-i+1)%MOD*inv2%MOD*cal(x/i)%MOD) %= MOD; 
    }
    return mp[x] = ans;
}
void work() {
    get_f(); inv2 = quick_pow(2, MOD-2), inv6 = quick_pow(6, MOD-2); read(n);
    writeln(1); writeln((cal(n)+MOD)%MOD);
}
int main() {
    work();
    return 0;
}