hdu 6053 TrickGCD 筛法

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

2017 Multi-University Training Contest - Team 2

 

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 ;
}