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) ]

GCD Expectation

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Time Limit: 4 Seconds      Memory Limit: 262144 KB

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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

15 11 2 3 4 5

Sample Output

42

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Author: LIN, Xi
Source: The 15th Zhejiang University Programming Contest
Submit    Status

/* ***********************************************Author        :CKbossCreated 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;}