[莫比乌斯函数][分段] hdu6053 TrickGCD (2017 Multi-University Training Contest - Team 2)

@(ACM题目)[莫比乌斯]

Description

You are given an array A , and Zhu wants to know there are how many different array B satisfy the following conditions?
* $1\leq B_i \leq A_i$
* For each pair( l , r ) ($1 \leq l \leq r \leq n$) , $gcd(bl,bl+1...br)\geq2$

Input

The first line is an integer T(1≤T≤10) describe the number of test cases.
Each test case begins with an integer number n describe the size of array A.
Then a line contains n numbers describe each element of A
You can assume that $1 \leq n, A_i \leq 10^5$

Output

For the kth test case , first output “Case #k: ” , then output an integer as answer in a single line . because the answer may be large , so you are only need to output answer mod $10^9+7$

Sample Input

1
4
4 4 4 4

Sample Output

Case #1: 17

题目分析

本题给定一个长度为$n$的数组$\{a_i\}$，现要构造一个新数组$\{b_i\}$，使其符合下列全部要求：
- $\{a_i\}$与$\{b_i\}$长度相同，且新数组对应元素小于等于原数组，即$b_i \leq a_i$
- 整个新数组的$\gcd \geq 2$

易得答案为：

$$\sum_{i = 2}^{\min(a)} - \mu(i) \prod_{j = 1}^{n} \lfloor \frac{a_j}{i} \rfloor$$
求解该式需要以下两方面：

莫比乌斯函数

$\mu(d)$为莫比乌斯函数，其定义如下：
- 若$d=1$，$\mu(d)=1$
- 若$d=p_1p_2\dots p_k$，$p_i$为互异素数，$\mu(d)=(-1)^k$
- 其他情况$\mu(d)=0$

我们可以莫比乌斯函数方便地进行容斥。

分段

对于上式中一个特定的$i$，我们可以将$\lfloor \frac{a_j}{i} \rfloor$为一个特定的值$k$的一段通过前缀和+快速幂快速地计算出来。
$k = \lfloor \frac{a_j}{i} \rfloor$说明$a_j$的值在$[ki, ki + i -1]$之间，用前缀和可以算出符合条件的$a_j$有$m$个，则其在$\prod_{j = 1}^{n} \lfloor \frac{a_j}{i} \rfloor$中的贡献为$k^m$，这个值可以用快速幂求得。
如果你蜜汁TLE了，请注意前缀和数组要开2e5而不是1e5，因为下标最大访问到$ki+i-1$

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 2e5+8;
const int M = 1e9+7;
bitset<maxn> isPrime;
int sum[maxn], primes[maxn], mu[maxn], tot = 0;
void mobius()
{
    isPrime.set();
    isPrime[1] = 0;
    mu[1] = 1;
    tot = 0;
    for(int i = 2; i < 100005; ++i)
    {
        if(isPrime.test(i)) primes[tot++] = i, mu[i] = -1;
        int d;
        for(int j = 0; j < tot && (d = i * primes[j]) < maxn; ++j)
        {
            isPrime[d] = false;
            if(i % primes[j]) mu[d] = -mu[i];
            else
            {
                mu[d] = 0;
                break;
            }
        }
    }
}

LL pow_m(LL a, LL n, LL M)
{
    a %= M;
    LL res = 1;
    while(n > 0)
    {
        if(n & 1) res = res*a%M;
        a = a * a % M;
        n >>= 1;
    }
    return res;
}

int main()
{
    mobius();
    int T, Case = 1;
    cin >> T;
    while(T--)
    {
        int n;
        scanf("%d", &n);
        int mn = 1e5+5, mx = -1;
        memset(sum, 0, sizeof sum);
        for(int i = 0; i < n; ++i)
        {
            int x;
            scanf("%d", &x);
            ++sum[x];
            if(mn > x) mn = x;
            if(mx < x) mx = x;
        }
        for(int i = 1; i < maxn; ++i) sum[i] += sum[i-1];
        LL res = 0;
        for(int i = 2; i <= mn; ++i)
        {
            if(!mu[i]) continue;
            LL tmp = 1;
            for(int j = 1; i*j <= mx; ++j)
                tmp = tmp * pow_m(j, sum[i*j + (i-1)] - sum[i*j - 1], M) % M;
            res = (res - mu[i] * tmp + M) % M;
        }
        printf("Case #%d: %lld\n", Case++, res);
    }
    return 0;
}