hdu 6053 莫比乌斯反演函数的利用 2017 Multi-University Training Contest

TrickGCD

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 622    Accepted Submission(s): 236

Problem 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 answermod109+7

 

Sample Input

144 4 4 4

 

Sample Output

Case #1: 17

 

Source

2017 Multi-University Training Contest - Team 2

题意：

给出 A数组，求有多少个数组满足 1≤Bi≤Ai  时 ，全体的gcd大于等于二

题解：

枚举gcd=k   范围是 2 ~ MIN  （MIN 表示A数组中最小的数字）

再枚举k的倍数

对于每个 k，在a[i]这个位置上有 a[i] / k 个数能满足条件构成 b[i]，只要把每个位置求出来累乘就好了

下面代码累乘的时候，得到了一个 指数，用快速幂处理即可

此时发现有重复现象

需要进行容斥搞一下就行了，然后容易发现使用莫比乌斯函数可以很快的解决

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;typedef long long ll;const int INF=0x3f3f3f3f;const int MAXN=100000+1000;const int mod=1e9+7;int cnt[MAXN];bool check[MAXN];int primer[MAXN];int mu[MAXN];void Moblus(){    mu[1]=1;    int tot=1;    for(int i=2;i<=MAXN;i++){        if(!check[i]){            primer[tot++]=i;            mu[i]=-1;        }        for(int j=1;j<tot&&i*primer[j]<=MAXN;j++){            check[i*primer[j]]=true;            if(i%primer[j]==0){                mu[i*primer[j]]=0;                break;            }            mu[i*primer[j]]=-mu[i];        }    }}ll qmi(ll a,ll b){    ll ans=1;    while(b)    {        if(b%2==1)ans=ans*a%mod;        a=(a*a)%mod;        b/=2;    }    return ans;}int main(){    Moblus();    freopen("in.txt","r",stdin);    int n,cases=0,x,T;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        int MIN=INF;        memset(cnt,0,sizeof(cnt));        for(int i=1;i<=n;i++){            scanf("%d",&x);            MIN=min(MIN,x);            cnt[x]++;        }        for(int i=1;i<MAXN;i++)            cnt[i]+=cnt[i-1];        ll ans=0,t;        for(int i=2;i<=MIN;i++){            t=1;            for(int j=1;i*j<=100000;j++)                t=t*qmi(j,cnt[(j+1)*i-1]-cnt[j*i-1])%mod;            ans=(ans-t*mu[i]%mod+mod)%mod;        }        printf("Case #%d: %lld\n",++cases,ans);    }    return 0;}