Min25筛求1-n内的素数和
作者：互联网
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 }
//#include <bits/stdc++.h> #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<string> #include<vector> #include<stack> #include<bitset> #include<cstdlib> #include<cmath> #include<set> #include<list> #include<deque> #include<map> #include<queue>
using namespace std;
const int N = 1000010;
typedef long long LL; LL TT,nn,k; namespace Min25 {
int prime[N], id1[N], id2[N], flag[N], ncnt, m; LL g[N], sum[N], a[N], T, n; inline void fff() { for(int i=0;i<=N;i++){ prime[i]=0; id1[i]=0; id2[i]=0; flag[i]=0; g[i]=0; sum[i]=0; a[i]=0; } ncnt=0; m=0; T=0; n=0; } inline int ID(LL x) { return x <= T ? id1[x] : id2[n / x]; }
inline LL calc(LL x) { return x * (x + 1) / 2 - 1; }
inline LL f(LL x) { return x; }
inline void init() { T = sqrt(n + 0.5); for (int i = 2; i <= T; i++) { if (!flag[i]) prime[++ncnt] = i, sum[ncnt] = sum[ncnt - 1] + i; for (int j = 1; j <= ncnt && i * prime[j] <= T; j++) { flag[i * prime[j]] = 1; if (i % prime[j] == 0) break; } } for (LL l = 1; l <= n; l = n / (n / l) + 1) { a[++m] = n / l; if (a[m] <= T) id1[a[m]] = m; else id2[n / a[m]] = m; g[m] = calc(a[m]); } for (int i = 1; i <= ncnt; i++) for (int j = 1; j <= m && (LL)prime[i] * prime[i] <= a[j]; j++) g[j] = g[j] - (LL)prime[i] * (g[ID(a[j] / prime[i])] - sum[i - 1]); }
inline LL solve(LL x) { if (x <= 1) return x; return n = x, init(), g[ID(n)]; }
}
void extend_gcd(LL a,LL b,LL &x,LL &y) { if(b==0) { x=1,y=0; return; } extend_gcd(b,a%b,x,y); LL tmp=x; x=y; y=tmp-(a/b)*y; } LL mod_inverse(LL a,LL m) { LL x,y; extend_gcd(a,m,x,y); return (m+x%m)%m; } /*int main() { LL sum; cin>>TT; while(TT--){ sum=0;
scanf("%lld%lld",&nn,&k); if(nn==1){ printf("0\n"); }else if(nn==2){ printf("6\n"); }else{ LL a=mod_inverse(2,k); //cout<<a<<endl; sum=(((((nn%k)*(nn%k))%k+(nn*3)%k)%k)*a)%k; //cout<<sum<<endl; Min25::fff(); sum=(sum-4+Min25::solve(nn+1))%k; printf("%lld\n",sum%k); } } return 0; }*/ int main() { int n; scanf("%d",&n); printf("%d\n",Min25::solve(n)); }
标签：nn,int,Min25,筛求,素数,sum,include,LL 来源： https://www.cnblogs.com/zpj61/p/13702729.html