【题目链接】
【算法】
设取的所有数都是k的约数,则这些数的lcm必然不大于k。
对于[1, m]中的每个数,统计a中有多少个数是它的约数即可。
【代码】
#include<bits/stdc++.h> using namespace std; typedef long long ll; const ll MAXN = 1e6; ll i,j,tmp,num,maxx,n,m,res; ll a[MAXN+10],sum[MAXN+10],h[MAXN+10]; vector<ll> ans; template <typename T> inline void read(T &x) { ll f = 1; x = 0; char c = getchar(); for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; } for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> inline void write(T x) { if (x < 0) { putchar('-'); x = -x; } if (x > 9) write(x/10); putchar(x%10+'0'); } template <typename T> inline void writeln(T x) { write(x); puts(""); } inline ll gcd(ll x,ll y) { return y == 0 ? x : gcd(y,x%y); } int main() { read(n); read(m); for (i = 1; i <= n; i++) { read(a[i]); if (a[i] <= m) ++h[a[i]]; } for (i = 1; i <= m; i++) { tmp = i; if (!h[i]) continue; for (j = tmp; j <= m; j += tmp) { sum[j] += h[i]; } } for (i = 1; i <= m; i++) { if (sum[i] > maxx) { maxx = sum[i]; num = i; } } if (!num) { printf("1 0 "); return 0; } for (i = 1; i <= n; i++) { if (!(num % a[i])) ans.push_back(i); } res = a[ans[0]]; for (i = 1; i < ans.size(); i++) res = res * a[ans[i]] / gcd(res,a[ans[i]]); write(res); putchar(' '); write(ans.size()); puts(""); for (i = 0; i < ans.size(); i++) { write(ans[i]); if (i < ans.size() - 1) putchar(' '); } return 0; }