Min25筛求素数和

复杂度n0.75/logn，实际上比n2/3的杜教筛要快一些。

  1 //#include <bits/stdc++.h>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<iostream>
  6 #include<string>
  7 #include<vector>
  8 #include<stack>
  9 #include<bitset>
 10 #include<cstdlib>
 11 #include<cmath>
 12 #include<set>
 13 #include<list>
 14 #include<deque>
 15 #include<map>
 16 #include<queue>
 17 
 18 using namespace std;
 19 
 20 const int N = 1000010;
 21 
 22 typedef long long LL;
 23 LL TT,nn,k;
 24 namespace Min25 {
 25 
 26     int prime[N], id1[N], id2[N], flag[N], ncnt, m;
 27 
 28     LL g[N], sum[N], a[N], T, n;
 29     inline void fff()
 30     {
 31         for(int i=0;i<=N;i++){
 32             prime[i]=0;
 33             id1[i]=0;
 34             id2[i]=0;
 35             flag[i]=0;
 36             g[i]=0;
 37             sum[i]=0;
 38             a[i]=0;
 39         }
 40         ncnt=0;
 41         m=0;
 42         T=0;
 43         n=0;
 44     }
 45     inline int ID(LL x) {
 46         return x <= T ? id1[x] : id2[n / x];
 47     }
 48 
 49     inline LL calc(LL x) {
 50         return x * (x + 1) / 2 - 1;
 51     }
 52 
 53     inline LL f(LL x) {
 54         return x;
 55     }
 56 
 57     inline void init() {
 58         T = sqrt(n + 0.5);
 59         for (int i = 2; i <= T; i++) {
 60             if (!flag[i]) prime[++ncnt] = i, sum[ncnt] = sum[ncnt - 1] + i;
 61             for (int j = 1; j <= ncnt && i * prime[j] <= T; j++) {
 62                 flag[i * prime[j]] = 1;
 63                 if (i % prime[j] == 0) break;
 64             }
 65         }
 66         for (LL l = 1; l <= n; l = n / (n / l) + 1) {
 67             a[++m] = n / l;
 68             if (a[m] <= T) id1[a[m]] = m; else id2[n / a[m]] = m;
 69             g[m] = calc(a[m]);
 70         }
 71         for (int i = 1; i <= ncnt; i++)
 72             for (int j = 1; j <= m && (LL)prime[i] * prime[i] <= a[j]; j++)
 73                 g[j] = g[j] - (LL)prime[i] * (g[ID(a[j] / prime[i])] - sum[i - 1]);
 74     }
 75 
 76     inline LL solve(LL x) {
 77         if (x <= 1) return x;
 78         return n = x, init(), g[ID(n)];
 79     }
 80 
 81 }
 82 
 83 void extend_gcd(LL a,LL b,LL &x,LL &y)
 84 {
 85     if(b==0) {
 86         x=1,y=0;
 87         return;
 88     }
 89     extend_gcd(b,a%b,x,y);
 90     LL tmp=x;
 91     x=y;
 92     y=tmp-(a/b)*y;
 93 }
 94 LL mod_inverse(LL a,LL m)
 95 {
 96     LL x,y;
 97     extend_gcd(a,m,x,y);
 98     return (m+x%m)%m;
 99 }
100 /*int main()
101 {
102     LL sum;
103     cin>>TT;
104     while(TT--){
105         sum=0;
106 
107         scanf("%lld%lld",&nn,&k);
108         if(nn==1){
109             printf("0\n");
110         }else if(nn==2){
111             printf("6\n");
112         }else{
113         LL a=mod_inverse(2,k);
114         //cout<<a<<endl;
115         sum=(((((nn%k)*(nn%k))%k+(nn*3)%k)%k)*a)%k;
116         //cout<<sum<<endl;
117         Min25::fff();
118         sum=(sum-4+Min25::solve(nn+1))%k;
119         printf("%lld\n",sum%k);
120         }
121     }
122     return 0;
123 }*/
124 int main()
125 {
126     int n;
127     scanf("%d",&n);
128     printf("%d\n",Min25::solve(n));
129 }

 

posted @ 2020-11-06 17:58  Lovaer  阅读(257)  评论(0)  编辑  收藏  举报