Longest Subsequence CodeForces - 632D (lcm)
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大意: 给定序列$a$, 求选出最长的一个子序列, 使得lcm不超过m.
刚开始想复杂了, 想着枚举gcd然后背包, 这样复杂度就是$O(\sum\limits_{i=1}^m \frac{m\sigma_0(i)}{i})$...... 估计了一下1e6大概只有1e8, 感觉剪个枝应该就可以过了, 打到最后才发现似乎不能输出方案...
看题解后发现就是个沙茶题, 直接枚举lcm即可.
#include <iostream> #include <random> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ‘\n‘ #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘ ‘;hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;} //head #ifdef ONLINE_JUDGE const int N = 1e6+10; #else const int N = 111; #endif int n, m, sum[N], a[N], f[N]; int main() { scanf("%d%d", &n, &m); REP(i,1,n) { scanf("%d", a+i); if (a[i]<=m) ++f[a[i]]; } REP(i,1,m) if (f[i]) { for (int j=i; j<=m; j+=i) sum[j]+=f[i]; } int ans = 0, pos = 0; REP(i,1,m) if (sum[i]>ans) ans=sum[i],pos=i; if (!pos) return puts("1 0"),0; printf("%d %d\n", pos, ans); REP(i,1,n) if (pos%a[i]==0) printf("%d ", i);hr; }
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