hdu 6053 TrickGCD 筛法

时间：2022-12-20 15:27:42

TrickGCD

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

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
 
Source
 
Recommend
liuyiding
题意：给你n个数字，每个位置的数字可以小于等于a[i]，求所有gcd(l,r)都满足大于等于2的情况数；
思路：显然枚举gcd的情况，对于每个位置都有a[i]/gcd的个数可以满足条件，gcd的情况的所有a[i]/gcd的乘积；
　　　这个也需要优化，枚举除数，a[i]/gcd相同的为一块，nlong（n)的复杂度*快速幂的log，后面的用容斥筛一下就好了；
　　　莫比乌斯好像也可以。
 
#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<iostream>

#include<cstdio>

#include<cmath>

#include<string>

#include<queue>

#include<algorithm>

#include<stack>

#include<cstring>

#include<vector>

#include<list>

#include<set>

#include<map>

#include<bitset>

#include<time.h>

using namespace std;

#define LL long long

#define pi (4*atan(1.0))

#define eps 1e-4

#define bug(x)  cout<<"bug"<<x<<endl;

const int N=1e5+,M=1e6+,inf=1e9+,MOD=1e9+;

const LL INF=1e18+,mod=1e9+;

int a[N],sum[N];

LL dp[N],num[N];

LL qpow(LL a,LL b,LL c)

{

    LL ans=;

    while(b)

    {

        if(b%)ans=(ans*a)%c;

        b>>=;

        a=(a*a)%mod;

    }

    return ans;

}

int main()

{

    int n;

    int T,cas=;

    scanf("%d",&T);

    while(T--)

    {

        memset(a,,sizeof(a));

        memset(sum,,sizeof(sum));

        scanf("%d",&n);

        for(int i=;i<=n;i++)

        {

            int x;

            scanf("%d",&x);

            a[x]++;

        }

        for(int i=;i<=;i++)

            sum[i]=sum[i-]+a[i];

        for(int i=;i<=;i++)//枚举除数

        {

            num[i]=1LL;

            for(int j=;j<=;j+=i)

            {

                int b;

                if(j+i->)b=sum[]-sum[j-];

                else if(j==) b=sum[j+i-];

                else b=sum[j+i-]-sum[j-];

                int a=j/i;

                if(a==&&b)num[i]=;

                else if(b)num[i]=(num[i]*qpow(a,b,mod))%mod;

            }

        }

        for(int i=;i>=;i--)

        {

            dp[i]=num[i];

            for(int j=i+i;j<=;j+=i)

                dp[i]-=dp[j],dp[i]=(dp[i]%mod+mod)%mod;

        }

        LL ans=;

        for(int i=;i<=;i++)

            ans+=dp[i],ans%=mod;

        printf("Case #%d: %lld\n",cas++,ans);

    }

    return ;

}

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