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

@(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≤Bi≤Ai
* For each pair( l , r ) (1≤l≤r≤n) , gcd(bl,bl+1...br)≥2

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≤n,Ai≤105

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 109+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{ai}，现要构造一个新数组{bi}，使其符合下列全部要求：
- {ai}与{bi}长度相同，且新数组对应元素小于等于原数组，即bi≤ai
- 整个新数组的gcd≥2

易得答案为：

∑i=2min(a)−μ(i)∏j=1n⌊aji⌋

求解该式需要以下两方面：

莫比乌斯函数

μ(d)为莫比乌斯函数，其定义如下：
- 若d=1，μ(d)=1
- 若d=p1p2…pk，pi为互异素数，μ(d)=(−1)k
- 其他情况μ(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⌊aji⌋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=⌊aji⌋说明aj的值在[ki,ki+i−1]之间，用前缀和可以算出符合条件的aj有m个，则其在∏nj=1⌊aji⌋中的贡献为km，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;}