HDU 6053 TrickGCD(莫比乌斯反演+前缀和)

TrickGCD

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 3226 Accepted Submission(s): 1206

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 answer mod 109+7

Sample Input
1
4
4 4 4 4

Sample Output
Case #1: 17

题意：给你一个A数组，求B数组的个数，B数组要满足这几个条件
Bi≤Ai
Gcd(B1,B2,…..,Bn)≥2
思路：易发现答案是枚举gcd从2开始到ai中最小的数的数量
那么我们可以从2开始枚举gcd，ai/gcd就是比ai小的数里面有几个数含有gcd为因子的数所以也就是
这个式子∑∏⌊aigcd⌋但直接算会有重复，这就引进了莫比乌斯函数，利用不同的gcd的值应该互质，也就和莫比乌斯函数一致了
∑−μ(d)∏⌊aigcd⌋
直接一个一个计算的话会超时，那么我们发现3-5除于3的值都是1，而且当其越大这个区间也就越大，我们可以利用前缀和的方法来减少时间

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>const int mod=1000000007;using namespace std;const int maxn=100005;int prime[maxn];int mu[maxn];bool vis[maxn];int cnt;int sum[maxn];void Init(){    int N=maxn;    memset(prime,0,sizeof(prime));    memset(mu,0,sizeof(mu));    memset(vis,0,sizeof(vis));    mu[1] = 1;    cnt = 0;    for(int i=2; i<N; i++){        if(!vis[i]){            prime[cnt++] = i;            mu[i] = -1;        }        for(int j=0; j<cnt&&i*prime[j]<N; j++){            vis[i*prime[j]] = 1;            if(i%prime[j]) mu[i*prime[j]] = -mu[i];            else{                mu[i*prime[j]] = 0;                break;            }        }    }}long long  quickmod(long long a,long long n){    long long ans=1;    a=a%mod;    while(n>0)    {        if(n%2==1)            ans=ans*a%mod;        n/=2;        a=a*a%mod;    }    return ans;}int main(){    Init();    int t,cas=1;    scanf("%d",&t);    while(t--)    {        memset(sum,0,sizeof(sum));        int n;        scanf("%d",&n);        int min1=mod;        for(int i=1;i<=n;i++)        {            int x;            scanf("%d",&x);            sum[x]++;            min1=min(min1,x);        }        for(int i=1;i<maxn;i++) sum[i]+=sum[i-1];        long long ans=0;        for(int i=2;i<=min1;i++)        {            long long temp=1;            for(int j=i;j<maxn;j+=i)                temp=temp*quickmod(j/i,sum[min(j+i-1,maxn-1)]-sum[j-1])%mod;            ans=(ans+temp*mu[i]*-1+mod)%mod;        }        printf("Case #%d: ",cas++);        printf("%lld\n",ans);    }    return 0;}