[杜教筛] HDU5608. function

最新推荐文章于 2019-08-30 15:36:50 发布

原创最新推荐文章于 2019-08-30 15:36:50 发布 · 374 阅读

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裸题

S (n) = \sum i = 1 n (i 2 - 3 i + 2) - \sum i = 2 n S (⌊ n i ⌋)

$S(n)=\sum_{i=1}^n (i^2-3i+2)-\sum_{i=2}^n S(\lfloor{n\over i}\rfloor)$

(f * 1) (n) = n 2 - 3 n + 2

$(f*1)(n)=n^2-3n+2$

f (n) = (f * ϵ) (n) = (f * μ * 1) (n) = \sum d | n (d 2 - 3 d + 2) μ (n d)

$f(n)=(f*\epsilon)(n)=(f*\mu*1)(n)=\sum_{d|n}(d^2-3d+2)\mu({n\over d})$

$O(n\ln n)$ 预处理后上杜教筛

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>

using namespace std;

const int P=1e9+7,inv=(P+1)/6;

map<int,int> S;

int t,n,f[1000010];
int p[1000010],mu[1000010];

inline void Pre(){
  mu[1]=1;
  for(int i=2;i<=1000000;i++){
    if(!p[i]) p[++*p]=i,mu[i]=-1;
    for(int j=1;j<=*p && 1LL*i*p[j]<=1000000;j++){
      p[i*p[j]]=1;
      if(i%p[j]) mu[i*p[j]]=-mu[i];
      else break;
    }
  }
  for(int i=1;i<=1000000;i++){
    int cur=(1LL*i*i-3*i+2)%P;
    for(int j=i;j<=1000000;j+=i)
      f[j]=(f[j]+1LL*cur*mu[j/i])%P;
  }
  for(int i=1;i<=1000000;i++)
    f[i]=(f[i]+f[i-1])%P;
}

int Sum(int n){
  if(n<=1000000) return f[n];
  if(S.count(n)) return S[n];
  int &ret=S[n];
  ret=(1LL*n*(n+1)%P*(2*n+1)%P*inv-3LL*n*(n+1)/2+2*n)%P;
  for(int i=2,j;i<=n;i=j+1){
    j=n/(n/i);
    ret=(ret-1LL*(j-i+1)*Sum(n/i))%P;
  }
  return ret;
}

int main(){
  freopen("1.in","r",stdin);
  freopen("1.out","w",stdout);
  Pre(); scanf("%d",&t);
  while(t--){
    scanf("%d",&n);
    printf("%d\n",(P+Sum(n))%P);
  }
}