codeforces 1017F

codeforces-1017F The Neutral Zone

这题内存限制只有16MB,考虑将2和3的倍数筛去，剩下的数只有1e8个，然后用bitset做标记数组，内存就只需要12MB.
标记的数字下标刚好是i/3.

#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof(a))
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
#define sc scanf
#define itn int
#define STR clock_t startTime = clock();
#define END clock_t endTime = clock();cout << double(endTime - startTime) / CLOCKS_PER_SEC *1000<< "ms" << endl;
using namespace std;
const int N=1e8+2;
const long long mod=1e9+7;
const long long mod2=998244353;
const int oo=0x7fffffff;
const int sup=0x80000000;
typedef long long ll;
typedef unsigned int ui;
template <typename it>void db(it *begin,it *end){while(begin!=end)cout<<(*begin++)<<" ";puts("");}
template <typename it>
string to_str(it n){string s="";while(n)s+=n%10+'0',n/=10;reverse(s.begin(),s.end());return s;}
template <typename it>int o(it a){cout<<a<<endl;return 0;}
inline ll mul_64(ll x,ll y,ll c){return (x*y-(ll)((long double)x/c*y)*c+c)%c;}
inline ll ksm(ll a,ll b,ll c){ll ans=1;for(;b;b>>=1,a=a*a%c)if(b&1)ans=ans*a%c;return ans;}
inline void exgcd(ll a,ll b,ll &x,ll &y){if(!b)x=1,y=0;else exgcd(b,a%b,y,x),y-=x*(a/b);}
bitset<N>bit;
int n,A,B,C,D;
ui cal(ui x){
    return ui(x*x*x*A+x*x*B+x*C+D);
}
ui num(int x){
    ll ans=0,m=n;
    while(m)ans+=m/=x;
    return ans;
}
int main(){
    cin>>n>>A>>B>>C>>D;
    if(n==1)return o(0),0;
    ui ans=cal(2)*num(2)+cal(3)*num(3);
    for(int i=5;i<=n;i++){
        if(i%2==0||i%3==0)continue;
        if(bit[i/3]==0){
            ans+=cal(i)*num(i);
            for(int j=i;j<=n;j+=i){
                if(j%2==0||j%3==0)continue;
                bit[j/3]=1;
            }
        }
    }
    o(ans);
}