思路:枚举lcm, 每个lcm的答案只能由他的因子获得,类似素数筛搞一下。
#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int,int>
#define piii pair<int, pair<int,int> > using namespace std; const int N = 1e6 + ;
const int M = + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-; int n, m, tot, a[N], cnt[N], id[N]; vector<int> v[N]; void init() {
for(int i = ; i <= m; i++) {
for(int j = i; j <= m; j+= i) {
v[j].push_back(i);
}
}
} int main() {
scanf("%d%d", &n, &m);
for(int i = ; i <= n; i++) {
int x; scanf("%d", &x);
if(x <= m) {
id[tot] = i;
a[tot++] = x;
}
} if(tot == ) {
printf("1 0\n");
} else {
init();
for(int i = ; i < tot; i++) {
cnt[a[i]]++;
} int mx = , ans = ;
for(int i = m; i >= ; i--) {
int ret = ;
for(int t : v[i]) {
ret += cnt[t];
}
if(ret >= mx) {
mx = ret;
ans = i;
}
} printf("%d %d\n", ans, mx);
for(int i = ; i < tot; i++) {
if(ans % a[i] == ) {
printf("%d ", id[i]);
}
}
puts("");
}
return ;
}
/*
*/