【51nod】1227 平均最小公倍数

题解

这个故事告诉们数论函数不要往分式上跑，你推不出来
好久没推式子了这么明显的转化我都忘了= =

首先(A(n) = frac{1}{n} sum_{i = 1}^{n} frac{i * n}{gcd(i,n)})
然后显然可以把n消掉
(A(n) = sum_{i = 1}^{n} frac{i}{gcd(i,n)})
改为枚举约数
(A(n) = sum_{d = 1}^{n} frac{1}{d}sum_{i = 1}^{n} i [gcd(i,n) == d])
(A(n) = sum_{d | n} sum_{i = 1}^{frac{n}{d}} i [gcd(i,frac{n}{d}) == 1])
有个欧拉函数的性质是，小于这个数的且与这个数互质的数的和是
(frac{n phi(n) + [n = 1]}{2}) 挺好理解的，因为一个与n互质的数p，n - p也和n互质
(frac{n phi(n) + [n = 1]}{2} = sum_{i = 1}^{n} i [gcd(i,n) == 1])
(A(n) = frac{1}{2} (sum_{d | n} frac{n}{d} phi(frac{n}{d}) + 1))

(F(n) = sum_{i = 1}^{n} A(i))
(F(n) = frac{1}{2} (sum_{i = 1}^{n} sum_{d | i} frac{i}{d} phi(frac{i}{d}) + n))
(F(n) = frac{1}{2} (sum_{i = 1}^{n} sum_{d | i} d phi(d) + n))
(F(n) = frac{1}{2} (sum_{i = 1}^{n} sum_{d = 1}^{frac{n}{i}} d phi(d) + n))

我们发现这个东西可以构造卷积啊
(sum_{d = 1}^{n} d phi(d))
卷上一个(Id(x) = x^{2})
那么我们就有
(sum_{i = 1}^{n} i^2 = sum_{i = 1}^{n} sum_{d | i}d phi(d) frac{i}{d})
(sum_{i = 1}^{n} i^2 = sum_{k = 1}^{n} k sum_{d = 1}^{frac{n}{k}}d phi(d))
那么就有
(S(n) = frac{n(n +1)(2n + 1)}{6} - sum_{i = 2}^{n} i S(lfloor frac{n}{i} floor))
然后用数论分块求(F(n))即可

题解

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <map>
//#define ivorysi
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define mo 974711
#define MAXN 1000000
#define RG register
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
	    while(c < '0' || c > '9') {
			if(c == '-') f = -1;
			c = getchar();
	    }
	    while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {putchar('-');x = -x;}
    if(x >= 10) {
		out(x / 10);
    }
    putchar('0' + x % 10);
}
int N;
const int MOD = 1000000007;
struct node {
	int x,v,next;
}E[100006];
int head[mo + 5],sumE,Inv2,Inv6;
int prime[MAXN + 5],tot,S[MAXN + 5],phi[MAXN + 5];
bool nonprime[MAXN + 5];
int inc(int a,int b) {
	a = a + b;
	if(a >= MOD) a -= MOD;
	return a;
}
int mul(int a,int b) {
	return 1LL * a * b % MOD;
}
void add(int u,int x,int v) {
	E[++sumE].x = x;E[sumE].v = v;E[sumE].next = head[u];
	head[u] = sumE;
}
void Insert(int x,int v) {
	add(x % mo,x,v);
}
int Query(int x) {
	int u = x % mo;
	for(int i = head[u] ; i ; i = E[i].next) {
		if(E[i].x == x) return E[i].v;
	}
	return -1;
}
int f(int x) {
	if(x <= MAXN) return S[x];
	int c = Query(x);
	if(c != -1) return c;
	int res = 0;
	for(int i = 2 ; i <= x ; ++i) {
		int r = x / (x / i);
		res = inc(res,1LL * (r - i + 1) * (r + i) / 2 % MOD * f(x / i) % MOD);
		i = r;
	}
	res = inc(1LL * x * (x + 1) % MOD * (2 * x + 1) % MOD * Inv6 % MOD,MOD - res);
	Insert(x,res);
	return res;
}
int calc(int x) {
	int res = 0;
	for(int i = 1 ; i <= x ; ++i) {
		int r = x / (x / i);
		res = inc(res,mul(r - i + 1,f(x / i)));
		i = r;
	}
	res = inc(res,x);
	res = mul(res,Inv2);
	return res;
}
void Solve() {
	phi[1] = 1;S[1] = 1;
	for(int i = 2 ; i <= MAXN ; ++i) {
		if(!nonprime[i]) {
			prime[++tot] = i;
			phi[i] = i - 1;
		}
		for(int j = 1 ; j <= tot ; ++j) {
			if(prime[j] > MAXN / i) break;
			nonprime[prime[j] * i] = 1;
			if(i % prime[j] == 0) {phi[i * prime[j]] = phi[i] * prime[j];break;}
			else phi[i * prime[j]] = phi[i] * (prime[j] - 1);
		}
		S[i] = inc(S[i - 1],mul(phi[i],i));
	}
	Inv2 = (MOD + 1) / 2;
	Inv6 = 1LL * Inv2 * (MOD + 1) / 3 % MOD; 
	int a,b;
	read(a);read(b);
	out(inc(calc(b),MOD - calc(a - 1)));enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
   	Solve();
    return 0;
}