[杜教筛] BZOJ 4916 神犇和蒟蒻

第一问玩玩你的吧
第二问直接杜教筛

S(n)=n∗(n+1)∗(2n+1)6−∑i=2ni∗S(⌊ni⌋)

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const int P=1e9+7;const int INV2=(P+1)/2;const int INV6=(P+1)/6;inline int sum(int x){ return (ll)x*(x+1)%P*INV2%P; }inline int sum2(int x){ return (ll)x*(x+1)%P*(2*x+1)%P*INV6%P; }inline int sum(int l,int r){ return (sum(r)+P-sum(l-1))%P; }inline int sum2(int l,int r){ return (sum2(r)+P-sum2(l-1))%P; }const int N=1000005;int maxn;int prime[N],vst[N],num;int phi[N];inline void Pre(int n){  maxn=n; phi[1]=1;  for (int i=2;i<=n;i++){    if (!vst[i]) prime[++num]=i,phi[i]=i-1;    for (int j=1;j<=num && (ll)i*prime[j]<=n;j++){      vst[i*prime[j]]=1;      if (i%prime[j]==0){    phi[i*prime[j]]=phi[i]*prime[j]; break;      }else    phi[i*prime[j]]=phi[i]*(prime[j]-1);    }  }  for (int i=1;i<=n;i++) phi[i]=(phi[i-1]+(ll)phi[i]*i%P)%P;}int n;int f[N];inline ll S(int n){  if (n<=maxn) return phi[n];  if (f[::n/n]!=-1) return f[::n/n];  ll ret=sum2(n);  int l,r;  for (l=2;(ll)l*l<=n;l++) ret+=P-(ll)S(n/l)*l%P;  for (ll t=n/l;l<=n;l=r+1,t--){    r=n/t;    ret+=P-(ll)S(t)*sum(l,r)%P;  }  return f[::n/n]=ret%P;}int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  scanf("%d",&n);  //int t=1; while ((ll)t*t<=n) t++; t--;  //Pre(t);  Pre(1000000);  memset(f,-1,sizeof(f));  printf("1\n%d\n",S(n));  return 0;}