由于有 个位置两数列不相同, 于是 使用 构造 ,
枚举 , 即假设 是已知量,
设 有 个位置不能整除 ,
由于的限制, 这 个位置需要改变,
即至少改变个位置, 若 , 说明无解 .
于是若存在答案, 则 ,
个位置的数要改变, 方案数为
剩下还需改变 个位置上的数, 方案数为
由于 为 , 所以要减去 是 倍数的方案
#include<bits/stdc++.h>
#define reg register
int read(){
char c;
int s = 0, flag = 1;
while((c=getchar()) && !isdigit(c))
if(c == '-'){ c = getchar(), flag = -1; break ; }
while(isdigit(c)) s = s*10 + c-'0', c = getchar();
return s * flag;
}
const int maxn = 300005;
const int mod = 1000000007;
int N;
int M;
int K;
int x;
int A[maxn];
int jc[maxn];
int rev[maxn];
int tmp[maxn];
int Ans[maxn];
int C(int n, int m){
int fz = jc[n];
int fm = 1ll*jc[n-m]*jc[m] % mod;
fz = 1ll*fz*rev[fm] % mod;
return fz;
}
int KSM(int a, int b){
int s = 1;
while(b){
if(b & 1) s = 1ll*s*a % mod;
a = 1ll*a*a % mod;
b >>= 1;
}
return s;
}
void Init(){
rev[1] = 1;
for(reg int i = 2; i < maxn; i ++) rev[i] = ((1ll*(-mod/i) * 1ll*rev[mod%i]) %mod + mod) % mod;
jc[0] = 1;
for(reg int i = 1; i < maxn; i ++) jc[i] = 1ll*jc[i-1]*i % mod;
}
int main(){
freopen("gcd.in", "r", stdin);
freopen("gcd.out", "w", stdout);
Init();
N = read(); M = read(); K = read();
for(reg int i = 1; i <= N; i ++) A[i] = read();
for(reg int d = 1; d <= M; d ++){
x = 0;
for(reg int i = 1; i <= N; i ++) x += ((A[i]%d) != 0);
if(x > K){ Ans[d] = 0; continue ; }
int tmp_1 = KSM(M/d, x), tmp_2 = C(N, N-x), tmp_3 = KSM(M/d-1, K-x);
tmp[d] = (1ll*tmp_1*tmp_2%mod)*1ll*tmp_3 % mod;
}
for(reg int d = 1; d <= M; d ++){
Ans[d] = tmp[d];
for(reg int i = 2; i*d <= M; i ++) Ans[d] = ((Ans[d] - tmp[i*d])%mod + mod) % mod;
printf("%d
", Ans[d]);
}
return 0;
}