hdu 5901 count prime & code vs 3223 素数密度

hdu5901题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5901

code vs 3223题目链接：http://codevs.cn/problem/3223/

思路:主要是用了一个Meisell-Lehmer算法模板，复杂度O（n^(2/3)）。讲道理，我不是很懂（瞎说什么大实话....），下面输出请自己改

 #include<bits/stdc++.h>  

 using namespace std;  

 typedef long long LL;
 const LL N = 5e6 + ;
 bool np[N];
 int prime[N], pi[N];  

 int getprime() {
     int cnt = ;
     np[] = np[] = true;
     pi[] = pi[] = ;
     for(int i = ; i < N; ++i) {
         if(!np[i]) prime[++cnt] = i;
         pi[i] = cnt;
         for(int j = ; j <= cnt && i * prime[j] < N; ++j) {
             np[i * prime[j]] = true;
             if(i % prime[j] == )   break;
         }
     }
     return cnt;
 }
 const int M = ;
 const int PM =  *  *  *  *  *  * ;
 int phi[PM + ][M + ], sz[M + ];
 void init() {
     getprime();
     sz[] = ;
     for(int i = ; i <= PM; ++i)  phi[i][] = i;
     for(int i = ; i <= M; ++i) {
         sz[i] = prime[i] * sz[i - ];
         for(int j = ; j <= PM; ++j) {
             phi[j][i] = phi[j][i - ] - phi[j / prime[i]][i - ];
         }
     }
 }
 int sqrt2(LL x) {
     LL r = (LL)sqrt(x - 0.1);
     while(r * r <= x)   ++r;
     return int(r - );
 }
 int sqrt3(LL x) {
     LL r = (LL)cbrt(x - 0.1);
     while(r * r * r <= x)   ++r;
     return int(r - );
 }
 LL getphi(LL x, int s) {
     if(s == )  return x;
     if(s <= M)  return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s];
     if(x <= prime[s]*prime[s])   return pi[x] - s + ;
     if(x <= prime[s]*prime[s]*prime[s] && x < N) {
         int s2x = pi[sqrt2(x)];
         LL ans = pi[x] - (s2x + s - ) * (s2x - s + ) / ;
         for(int i = s + ; i <= s2x; ++i) {
             ans += pi[x / prime[i]];
         }
         return ans;
     }
     return getphi(x, s - ) - getphi(x / prime[s], s - );
 }
 LL getpi(LL x) {
     if(x < N)   return pi[x];
     LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - ;
     for(int i = pi[sqrt3(x)] + , ed = pi[sqrt2(x)]; i <= ed; ++i) {
         ans -= getpi(x / prime[i]) - i + ;
     }
     return ans;
 }
 LL lehmer_pi(LL x) {
     if(x < N)   return pi[x];
     int a = (int)lehmer_pi(sqrt2(sqrt2(x)));
     int b = (int)lehmer_pi(sqrt2(x));
     int c = (int)lehmer_pi(sqrt3(x));
     LL sum = getphi(x, a) + LL(b + a - ) * (b - a + ) / ;
     for (int i = a + ; i <= b; i++) {
         LL w = x / prime[i];
         sum -= lehmer_pi(w);
         if (i > c) continue;
         LL lim = lehmer_pi(sqrt2(w));
         for (int j = i; j <= lim; j++) {
             sum -= lehmer_pi(w / prime[j]) - (j - );
         }
     }
     return sum;
 }  

 int main() {
     init();
     LL n,m;
     while(cin >> m >> n)
     {
         cout<<prime[n]<<endl; //输出第n个素数
         cout<<pi[m]<<endl;         //输出该数是第几个素数
         cout << lehmer_pi(n) <<endl;    //输出区间[1,n]包含的素数
         cout << lehmer_pi(n)-lehmer_pi(m) << endl;    //判断区间(m,n]包含的素数 

             cout << lehmer_pi(n)-lehmer_pi(m-1) << endl;   //输出区间[m,n]包含的素数个数 

     }
     return ;
 } 

还有一个O(n^(4/3))算法，我又不懂，菜的抠脚....

 #include <bits/stdc++.h>
 #define ll long long
 using namespace std;
 ll f[],g[],n;
 void init(){
     ll i,j,m;
     for(m=;m*m<=n;++m)f[m]=n/m-;
     for(i=;i<=m;++i)g[i]=i-;
     for(i=;i<=m;++i){
         if(g[i]==g[i-])continue;
         for(j=;j<=min(m-,n/i/i);++j){
             if(i*j<m)f[j]-=f[i*j]-g[i-];
             else f[j]-=g[n/i/j]-g[i-];
         }
         for(j=m;j>=i*i;--j)g[j]-=g[j/i]-g[i-];
     }
 }
 int main(){
     while(scanf("%I64d",&n)!=EOF){
         init();
         cout<<f[]<<endl;
     }
     return ;
 }