POj3292 Semi-prime H-numbers

POj3292 Semi-prime H-numbers

　　Ed Screening + Violence Watch

first hit the H-prime table , each to find a h-prime, to sieve away its multiples,

Here can be pruned, only need to sift out similar 4n+1 such as the number, if I is h-prime, only need to sift off 5i,9i,... (4k+1) I, (k=1,2,3,4,5 ...)

The proof is as follows, set X*I is a multiple of I, just make x*i=4m+1, (m=1,2,3,...) When you find the X

Set I=4n+1, then:

(x*i)%4=x* (4*n+1)%4= (x*4*n+x)%4=x%4=1;

So X=4k+1 (k=1,2,3,4 ...)

　　h-prime table, and then hit the H-semi-primes table , each enumeration two h-prime, get H-semi-primes, and Mark H-semi-primes.

After the table, 16ms water over, because the second table when the judge condition is (h_prime[i]*h_prime[i]<=size), so the final complexity is still smaller than O (n^2).

/** created:2016 March 30 20:57 49 Second Wednesday * author:akrusher**/#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<ctime>#include<iostream>#include<algorithm>#include<string>#include<vector>#include<deque>#include<list>#include<Set>#include<map>#include<stack>#include<queue>#include<numeric>#include<iomanip>#include<bitset>#include<sstream>#include<fstream>using namespacestd;#defineRep (i,a,n) for (int i=a;i<n;i++)#definePer (i,a,n) for (int i=n-1;i>=a;i--)#defineIn (n) scanf ("%d",& (n))#defineIn2 (X1,X2) scanf ("%d%d",& (x1),& (x2))#defineINLL (n) scanf ("%i64d",& (n))#defineInll2 (X1,X2) scanf ("%i64d%i64d",& (x1),& (x2))#defineINLLD (n) scanf ("%lld",& (n))#defineInlld2 (X1,X2) scanf ("%lld%lld",& (x1),& (x2))#defineINF (n) scanf ("%f",& (n))#defineInf2 (X1,X2) scanf ("%f%f",& (x1),& (x2))#defineINLF (n) scanf ("%lf",& (n))#defineINLF2 (X1,X2) scanf ("%lf%lf",& (x1),& (x2))#defineInc (STR) scanf ("%c",& (str))#defineINS (str) scanf ("%s", (str))#defineOut (x) printf ("%d\n", (x))#defineOut2 (x1,x2) printf ("%d%d\n", (x1), (x2))#defineOutf (x) printf ("%f\n", (x))#defineOUTLF (x) printf ("%lf\n", (x))#defineOUTLF2 (x1,x2) printf ("%lf%lf\n", (x1), (x2));#defineOUTLL (x) printf ("%i64d\n", (x))#defineOUTLLD (x) printf ("%lld\n", (x))#defineOUTC (str) printf ("%c\n", (str))#definePB Push_back#defineMP Make_pair#defineFi first#defineSe Second#defineSZ (x) ((int) (x). Size ())#defineMem (x, y) memset (x,y,sizeof (×));typedef vector<int>Vec;typedefLong LongLl;typedef pair<int,int>P;Const intdx[4]={1,0,-1,0},dy[4]={0,1,0,-1};Const intinf=0x3f3f3f3f;Constll mod=1e9+7; ll Powmod (ll A,ll b) {ll res=1; a%=mod; for(; b;b>>=1){if(b&1) Res=res*a%mod;a=a*a%mod;}returnRes;}Const BOOLAc=true; ll h_prime[1000005];//Storage H_primeBOOLis_h_prime[1000005];ll num[1000005];//count the number of products consisting of two h_prime number within IBOOLused[1000005];Constll size=1000001;//violence to the table, first hit the prime table, and then hit by two prime numbers of the tablevoidSieve () {//H_prime Filteringll q=0;  for(LL i=5; i<=size;i+=4) is_h_prime[i]=true;  for(LL i=5; i<=size;i+=4){        if(Is_h_prime[i]) {h_prime[q++]=i;  for(LL j=5*i;j<=size;j+=4*i) is_h_prime[j]=false;//filter out multiple + pruning (filter 4n+1 only)        }        }}voidFactor () {//enumeration of the number of products consisting of two h_prime numbers (whether the tag is used)ll cnt=0;  for(LL i=0; h_prime[i]*h_prime[i]<=size;i++){         for(LL j=i;h_prime[j]*h_prime[i]<=size;j++){            if(!used[h_prime[j]*H_prime[i]]) {Used[h_prime[j]*h_prime[i]]=true; }        }    }     for(LL i=1; i<=size;i++) {//H-semi-primes playing table    if(Used[i]) {CNT++; } Num[i]=cnt;//num has to be assigned regardless of the CNT addition. }}            intMain () {ll n;    Sieve (); factor ();  while(INLLD (n) = =1){    if(n==0) Break; printf ("%lld%lld\n", N,num[n]); }    return 0;}

POj3292 Semi-prime H-numbers