HDU 6706 huntian oy

$HDU\ \ 6706\ \ huntian\ oy$

http://acm.hdu.edu.cn/showproblem.php?pid=6706

哎，開始就覺得,對於$,i>j,gcd(i,j)=1$

$gcd(i^a-j^a,i^b-j^b)=i^{gcd(a,b)}-j^{gcd(a,b)}$

手殘，好久沒刷題了 ，忘了很多東西 哈哈
藉着題目又複習了一下。
其實對於：

$(i^a-j^a)=(i-j)\sum_{k=0}^{a-1}i^{a-1}\Big(\frac{j}{i}\Big)^k$

$令b>a, b=ta+r$

$(i^b-j^b)=(i^{ta+r}-j^{ta+r})=(i-j)\sum_{k=0}^{ta+r-1}i^{ta+r-1}\Big(\frac{j}{i}\Big)^k\\ =(i-j)\sum_{d=0}^{t-1}i^{^{(t-1-d)a+r}}j^{da}\sum_{k=0}^{a-1}i^{a-1}\Big(\frac{j}{i}\Big)^k+(i-j)j^{ta}\sum_{k=0}^{r-1}i^{r-1}\Big(\frac{j}{i}\Big)^k\\=\Big\{\sum_{d=0}^{t-1}i^{^{(t-1-d)a+r}}j^{da}\Big\}(i^a-j^a)+j^{ta}(i^r-j^r)$

又因爲$gcd(i,j)=1$.

$(i^b-j^b)\%\ (i^a-j^a)=(i^{b\%a}-j^{b\%a})$

所以：

$ans = \sum_{i=1}^{n}\sum_{j=1}^i(i-j)[gcd(i,j)=1]$

$ans = \sum_{i=1}^ni\varphi(i)-\sum_{i=1}^n\frac{1+i\varphi(i)}{2}\\=\sum_{i=2}^n\frac{i\varphi(i)}{2}$

這種計算方便統計小於$10^6$的答案。

自己用了一個很蠢的方法。。。。。。。。

如果前面使用莫比烏斯反演的話有：

令：$F(n)=\sum_{i=1}^{n}\sum_{j=1}^i(i-j)$

$ans =\sum_{k=1}^n\mu(k)k\sum_{i=1}^{\lfloor\frac{n}{k}\rfloor}\sum_{j=1}^i(i-j)=\sum_{k=1}^n\mu(k)kF(\Big\lfloor\frac{n}{k}\Big\rfloor)$

這個姿態有點慢。需要用前面的方法打標。後面這種方法需要計算 $mu(k)k$前綴和$M$，然後查表。

$ans = \sum_{L=1}^{m}F(L)(M(\Big\lfloor\frac{n}{L}\Big\rfloor-M(\Big\lfloor\frac{n}{L+1}\Big\rfloor))+\sum_{k=1}^{\lfloor\frac{n}{m+1}\rfloor}\mu(k)kF(\Big\lfloor\frac{n}{k}\Big\rfloor)$

#pragma GCC diagnostic error "-std=c++11"
#pragma GCC target("avx")
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#pragma GCC optimize(3)
#include <stdio.h>
#include <cmath>
#define MAXN 1000006
using namespace std;
typedef long long LL;

int mod = 1e9 + 7;
int mu[MAXN + 5] = { 0,-1 };
int M[MAXN + 5];
int tmp[MAXN + 5];
int In[100];

inline int Sn(int n)
{
	if (n & 1)
		return (LL)n*((n + 1) >> 1) % mod;
	else
		return (LL)(n >> 1)*(n + 1) % mod;
}

void clat(int n, int d)
{
	int ans = 0;
	int m = (int)sqrt(n) + 1;
	for (int L = 1; L < m; L++)
	{
		ans += (LL)M[L] * (Sn(n / L) - Sn(n / (L + 1)) + mod) % mod;
		if (ans >= mod)ans -= mod;
	}
	for (int i = (int)(n / m); i > 1; i--)
	{
		int u = n / i;
		if (u < MAXN)
			ans += (LL)i*M[u] % mod;
		else
			ans += (LL)i*tmp[i*d] % mod;
		if (ans >= mod)ans -= mod;
	}
	tmp[d] = (1 - ans + mod) % mod;
}

int slove(int n)
{
	if (n < MAXN)return M[n];
	for (int i = (n / (MAXN - 1)); i; i--) clat(n / i, i);
	return tmp[1];
}

int phi[MAXN + 10];

void init()
{
	for (int i = 1; i <= MAXN; i++)  phi[i] = i;
	for (int i = 2; i <= MAXN; i += 2) phi[i] /= 2;
	for (int i = 3; i <= MAXN; i += 2)
		if (phi[i] == i) {
			for (int j = i; j <= MAXN; j += i) {
				phi[j] = phi[j] / i*(i - 1);
			}
		}
}

inline int F(int n)
{
	LL tm = LL(n)*(n + 1) % mod;
	int ret = (tm*(2 * n + 1) % mod)*In[12] % mod;
	ret -= tm*In[4] % mod;
	if (ret < 0)ret += mod;
	return ret;
}

int S[MAXN + 5];

inline int getM(int u, int d)// u = n/d
{
	if (u < MAXN)return M[u];
	return tmp[d];
}

int main()
{
	
	init();
	for (int i = 1; i < MAXN; i++)
	{
		mu[i] = -mu[i];
		for (int j = (i << 1); j < MAXN; j += i)mu[j] += mu[i];
		if (mu[i] < 0)mu[i] += mod;
		M[i] = M[i - 1] + LL(mu[i]) * i%mod;
		if (M[i] >= mod)M[i] -= mod;
	}
	In[1] = 1;
	for (int i = 2; i < 100; i++)
	{
		In[i] = LL(mod - (mod / i))*In[mod%i] % mod;
	}
	for (int i = 2; i < MAXN; i++)
	{
		S[i] = S[i - 1] + ((LL)phi[i] * i%mod)*In[2] % mod;
		if (S[i] >= mod)S[i] -= mod;
	}

	int T = 10;

	scanf("%d", &T);

	while (T--)
	{
		int n, a, b;
		scanf("%d %d %d", &n, &a, &b);
		if (n < MAXN)
		{
			printf("%d\n", S[n]);
			continue;
		}
		slove(n);
		int m = sqrt(n) + 1;
		int ans = 0;
		for (int L = 1; L < m; L++)
		{
			int mm = getM(n / L, L) - getM(n / (L + 1), L + 1);
			if (mm < 0)mm += mod;
			ans += (LL)F(L)*mm%mod;
			if (ans >= mod)ans %= mod;
		}
		int size = n / m;
		for (int k = 1; k <= size; k++)
		{
			ans += ((LL)k*mu[k] % mod) * F(n / k) % mod;
			if (ans >= mod)ans -= mod;
		}
		printf("%d\n", ans);
	}
	return 0;
}