2017ACM暑期多校联合训练 - Team 8 1002 HDU 6134 Battlestation Operational （数论 莫比乌斯反演）

题目链接

Problem Description

The Death Star, known officially as the DS-1 Orbital Battle Station, also known as the Death Star I, the First Death Star, Project Stardust internally, and simply the Ultimate Weapon in early development stages, was a moon-sized, deep-space mobile battle station constructed by the Galactic Empire. Designed to fire a single planet-destroying superlaser powered by massive kyber crystals, it was the pet project of the Emperor, Darth Vader, and its eventual commander Grand Moff Wilhuff Tarkin to expound the military philosophy of the aptly named Tarkin Doctrine.

— Wookieepedia

In the story of the Rogue One, the rebels risked their lives stolen the construction plan of the Death Star before it can cause catastrophic damage to the rebel base. According to the documents, the main weapon of the Death Star, the Superlaser, emits asymmetric energy in the battlefield that cause photons to annihilate and burns everything in a single shot.

You are assigned the task to estimate the damage of one shot of the Superlaser.

Assuming that the battlefield is an n×n grid. The energy field ignited by the Superlaser is asymmetric over the grid. For the cell at i-th row and j-th column, ⌈i/j⌉ units of damage will be caused. Furthermore, due to the quantum effects, the energies in a cell cancel out if gcd(i,j)≠1 or i<j.

The figure below illustrates the damage caused to each cell for n=100. A cell in black indicates that this cell will not be damaged due to the quantum effects. Otherwise, different colors denote different units of damages.

Your should calculate the total damage to the battlefield. Formally, you should compute

where [(i,j)=1] evaluates to be 1 if gcd(i,j)=1, otherwise 0.

Input

There are multiple test cases.

Each line of the input, there is an integer n (1≤n≤106), as described in the problem.

There are up to 104 test cases.

Output

For each test case, output one integer in one line denoting the total damage of the Superlaser, f(n) mod 109+7.

Sample Input

1

2

3

10

Sample Output

1

3

8

110

题意：

当i于j不同时，ceil(i/j)的值可能取到一样或者不一样的值，不同的值代表着不同的颜色，求出在一个n*n的矩形的范围内，这样的值的和是多少。

分析：

已知：f(n) = sigma[1<=i<=n]sigma[1<=j<=i]ceil[i/j] (gcd(i,j)==1)

给定一个n，求f(n);

可以推出公式： n = sigma[d|n]phi[d] = sigma[d|n]phi[n/d];

phi[x]表示<=x的数与x互质的个数。

看一下这个公式的证明：

如果gcd(i,n)==d，那么可以得到：gcd(i/d,n/d)=1;

那么和n的最大公约数为d的个数为phi[n/d]; 所以n = sigma[d|n]phi[n/d] = sigma[d|n]phi[d];

根据n = sigma[d|n]phi[d];

那么有定义：

h(i)表示sigma[1<=j<=i]ceil[i/j] (gcd(i,j)==1) 这里的j都是和i互质时候计算的结果。

g(i)表示sigma[1<=j<=i]ceil[i/j]（不用考虑i于j是否互质）

那么h(i) = sigma[d|i]mu[d]*g(i/d);

计算所有的g(d)(1<=d<=n)通过枚举j跳在d中跳的方式处理出来，然后前缀和（也可以直接计算出来。不需要再求前缀和，具体看代码）

计算出来所有的h(i)。f(i) = sigma[1<=j<=i]h(j);

说的可能不太好理解，具体的求解过程可以看一下这个

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int N = 1e6+10;
const int mod = 1e9 + 7;
LL f[N], g[N], h[N];
int prime[N], tot, not_prime[N];
int mu[N];
void mobius()///求出一个数字对应的莫比乌斯函数
{
    mu[1] = 1;
    tot = 0;
    for(int i = 2; i < N; i++)
    {
        if(!not_prime[i])///如果当前的数字是质数，对应的莫比乌斯函数为-1
        {
            prime[++tot] = i;
            mu[i] = -1;
        }
        for(int j = 1; prime[j]*i<N; j++)
        {
            not_prime[prime[j]*i] = 1;///prime[j]*i可以是单个的质数相乘的形式（不存在两个及以上相同的质数相乘）
            if(i%prime[j]==0)///如果能将i用质数分解的话
            {
                mu[prime[j]*i] = 0;
                break;
            }
            mu[prime[j]*i] = -mu[i];
        }
    }
}

void init()
{
    for(int i = 1; i < N; i++)
    {
        g[i]++;
        for(int j = i+1; j < N; j+=i)
        {
            g[j]++;
        }
    }
    for(int i = 1; i < N; i++) g[i] = (g[i]+g[i-1])%mod;

    for(int i = 1; i < N; i++)
    {
        for(int j = i; j < N; j+=i)
        {
            h[j] = (h[j]+mu[i]*g[j/i]%mod+mod)%mod;
        }
    }
    for(int i = 1; i < N; i++)
    {
        f[i] = (f[i-1]+h[i])%mod;
    }
}
int main()
{
    int n;
    mobius();
    init();
    while(scanf("%d",&n)==1)
    {
        printf("%lld\n",f[n]);
    }
    return 0;
}

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