zoj 3868 GCD Expectation（容斥原理）
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
42
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. （(2n - 1) 其实不用考虑了）
Sample Input
1 5 1 1 2 3 4 5
Sample Output
42
这题一不小心就要超时了。。。
#include<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<vector>
#include<cmath>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#include <ctime>
#include <cstdlib>
#define ll long long
#define mod 998244353
using namespace std;
int num[1000001]; //把这两个数组放到main函数里就超时了。。
int r[1000001];
ll power(ll a, ll n) {
ll r = 1;
for (; n; n >>= 1) {
if (n & 1) r = r * a % mod;
a = a * a % mod;
}
return r;
}
int main(){
int t;
scanf("%d",&t);
while(t--){
memset(r,0,sizeof(r)); //我把r换成map居然超时了。。
int n,m,a;
scanf("%d%d",&n,&m);
int max=0;
for(int i=0;i<n;++i){
scanf("%d",&a);
r[a]++;
if(a>max)
max=a;
}
for(int i=max;i>=1;--i){ //枚举gcd的可能值(倒序)
int sum=0;
num[i]=0;
for(int j=i;j<=max;j+=i){ //求它的倍数
sum+=r[j]; //符合条件的元素个数
num[i]=(num[i]-num[j]+mod)%mod; //我们求的只有gcd=i而不包括它的倍数
}
num[i]=(num[i]+power(2,sum)-1)%mod; //power(2,sum)-1求长度为sum的非空子集个数
}
int s=0;
for(int i=1;i<=max;++i){ //gcd
s=(s+power(i,m)*num[i]%mod)%mod;
}
printf("%d\n",s);
}
return 0;
}