ZOJ3868:GCD Expectation

标签：fzu

Edward has a set of n integers {a₁, a₂,...,a_n}. He randomly picks a nonempty subset {x₁, x₂,…,x_m} (each nonempty subset has equal probability to be picked), and would like to know the expectation of [gcd(x₁, x₂,…,x_m)]^k.

Note that gcd(x₁, x₂,…,x_m) is the greatest common divisor of {x₁, x₂,…,x_m}.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains two integers n, k (1 ≤ n, k ≤ 10⁶). The second line contains n integers a₁, a₂,…,a_n (1 ≤ a_i ≤ 10⁶).

The sum of values max{a_i} for all the test cases does not exceed 2000000.

Output

For each case, if the expectation is E, output a single integer denotes E · (2ⁿ - 1) modulo 998244353.

Sample Input

1
5 1
1 2 3 4 5

Sample Output

42

对于N个数的序列，所有非空子集中，其期望是GCD的k次方

输出期望乘以（2^N-1）的值

题目中1的概率是26/31，2的概率是2/32,3,4,5的概率是1/32

期望则是42/32,所以答案为42，也就是说我们的目标是求出期望的分子部分即可

对于N的序列，肯定有2^N-1个非空子集，其中其最大的GCD不会大于原序列的max，那么我们用数组fun来记录其期望

例如题目中的，期望为1的有26个，期望为2的有2个，期望为3,4,5的都只有1个

我们可以拆分来算，首先对于1，期望为1,1的倍数有5个，那么这五个的全部非空子集为2^5-1种，得到S=(2^5-1)*1;

对于2,2的期望应该是2，但是在期望为1的时候所有的子集中，我们重复计算了2的期望，多以我们应该减去重复计算的期望数，现在2的期望应该作1算，那么对于2的倍数，有两个，2,4，其组成的非空子集有2^2-1个，所以得到S+=(2^2-1)*1

对于3,4,5同理；

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <algorithm>
using namespace std;
#define ls 2*i
#define rs 2*i+1
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,x) memset(a,x,sizeof(a))
#define w(a) while(a)
#define LL long long
const double pi = acos(-1.0);
#define Len 1000005
#define mod 998244353
const LL inf = 1<<30;

LL t,n,k;
LL a[Len];
LL two[Len],fun[Len],cnt[Len],vis[Len],maxn;
LL power(LL x, LL y)
{
    LL ans = 1;
    w(y)
    {
        if(y&1)
            ans=(ans*x)%mod;
        x=(x*x)%mod;
        y/=2;
    }
    return ans;
}
int main()
{
    LL i,j;
    scanf("%lld",&t);
    two[0] = 1;
    up(i,1,Len-1)
    two[i] = (two[i-1]*2)%mod;
    w(t--)
    {
        mem(cnt,0);
        mem(vis,0);
        scanf("%lld%lld",&n,&k);
        maxn = 0;
        up(i,0,n-1)
        {
            scanf("%lld",&a[i]);
            if(!vis[a[i]])
            {
                vis[a[i]] = 1;
                cnt[a[i]] = 1;
            }
            else
                cnt[a[i]]++;
            maxn = max(maxn,a[i]);
        }
        fun[1] = 1;
        up(i,2,maxn)
        fun[i] = power(i,k);
        up(i,1,maxn)
        {
            for(j = i+i; j<=maxn; j+=i)
                fun[j]=(fun[j]-fun[i])%mod;
        }
        LL ans = (two[n]-1)*fun[1]%mod;
        up(i,2,maxn)
        {
            LL cc = 0;
            for(j = i; j<=maxn; j+=i)
            {
                if(vis[j]) cc+=cnt[j];
            }
            LL tem = (two[cc]-1)*fun[i]%mod;
            ans = (ans+tem)%mod;
        }
        printf("%lld\n",(ans+mod)%mod);
    }

    return 0;
}

ZOJ3868:GCD Expectation

标签：fzu

原文地址：http://blog.csdn.net/libin56842/article/details/45016299