ZOJ 3868 GCD Expectation DP
dp[i] 表示公约数为i时有多少种组合
先预处理一遍dp[i]这是的dp[i]表示含有公约数i或者i的倍数的组合有多少个
再倒着dp dp[i] - = Sigma(dp[j]) (j是i的倍数 2i,3i,4i.....)
结果既为 Sigma[ dp[i]*pow(i,k) ]
Edward has a set of n integers {a1, a2,...,an}. He randomly picks a nonempty subset {x1, x2,…,xm} (each nonempty subset has equal probability to be picked), and would like to know the expectation of [gcd(x1, x2,…,xm)]k.
Note that gcd(x1, x2,…,xm) is the greatest common divisor of {x1, x2,…,xm}.
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,…,an (1 ≤ ai ≤ 106).
The sum of values max{ai} 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
42Author: LIN, Xi
Source: The 15th Zhejiang University Programming Contest
Submit Status
/* ***********************************************
Author :CKboss
Created Time :2015年04月13日 星期一 09时29分19秒
File Name :I.cpp
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long int LL;
const LL mod=998244353LL;
const int maxn=2001000;
LL power(LL x,int n)
{
LL e=1LL;
while(n) { if(n&1) e=(e*x)%mod; x=(x*x)%mod; n/=2; }
return e%mod;
}
int n,k,mx;
int vis[maxn];
LL two[maxn],dp[maxn];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
two[0]=1LL;
for(int i=1;i<maxn;i++) { two[i]=(two[i-1]*2LL)%mod; }
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d%d",&n,&k);
memset(vis,0,sizeof(vis)); memset(dp,0,sizeof(dp)); mx=0;
for(int i=0;i<n;i++) { int x; scanf("%d",&x); mx=max(mx,x); vis[x]++; }
dp[1]=(two[n]-1+mod)%mod;
for(int i=2;i<=mx;i++)
{
int temp=0;
for(int j=i;j<=mx;j+=i) if(vis[j]) temp+=vis[j];
dp[i]=(two[temp]-1+mod)%mod;
}
for(int i=mx;i>=1;i--)
{
for(int j=i+i;j<=mx;j+=i)
dp[i]=(dp[i]-dp[j]+mod)%mod;
}
LL sum=0;
for(int i=1;i<=mx;i++)
sum=(sum+(dp[i]*power(i,k))%mod)%mod;
cout<<sum<<endl;
}
return 0;
}