The number of prime numbers by Euler sieve method

Determine whether a is a prime number, and find 1--n of prime numbers

Consider the Euler sieve method ————

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//Determines whether the number is a prime number, or 1 to n#include <iostream>#include<cstring>#defineMAX 10000000using namespacestd;BOOLb[max+1];//b 0 is prime, 1 is non-primeintc[max+1];intSushu (intN) {memset (b,0, n +1); b[0]=1; b[1]=1;//1 non-prime and non-composite    intI, j, cnt =0, T;  for(i =2; I <= N; i++){        if(!B[i]) {c[cnt++] =i; }         for(j =0; j<cnt; J + +) {T= I*c[j];//prime numbers multiply prime numbers for composite, prime numbers multiply composite for composite, a composite must be written as prime numbers in the form of a number            if(T >N) Break; B[t]=true; if(I%c[j] = =0)                 Break; }    }    returnCNT;}intMain () {intT,a,n,c; CIN>>u;  while(t--) {cin>>a>>N; C=Sushu (n); if(b[a]==0) cout<<"YES"<<c<<Endl; Elsecout<<"NO"<<c<<Endl; }    return 0;}

Look at someone else's code for 1 billion seconds, and

#include <iostream>#include<cstring>#include<cstdio>#include<ctime>#include<algorithm>using namespacestd;intn=11000000;BOOLvisit[11000000];intprime[11000000];voidInit_prim () {memset (visit,true,sizeof(visit)); visit[0] = visit[1] =false; intnum =0;  for(inti =2; I <= N; ++i) {if(Visit[i] = =true) {num++; Prime[num]=i; }        /*For (int j = 1;  ((J <= num) && (i * prime[j] <= n));            ++J) {Visit[i * Prime[j]] = false;        if (i% prime[j] = = 0) break; }*/         for(intj =2; I *j <=n; J + +) {Visit[i* j] =false; }    }}intOutnum (intN) {    inti;  for(i=1;p rime[i]<=n;i++); returnI-1;}intMain () {clock_t start, finish; Doubleduration; Start=clock ();    Init_prim (); intT; CIN>>T;  while(t--)    {        intX,n; CIN>> x >>N; if(Visit[x]) cout<<"YES"<<" "; Elsecout<<"NO"<<" "; cout<< outnum (n) <<Endl; } Finish=clock (); Duration= (Double) (Finish-start)/clocks_per_sec*1000.0; printf ("%f ms\n", duration); return 0;} 

The number of prime numbers by Euler sieve method