"NOIP simulation Question" [T1] prime number generator (linear sieve + promiscuity)

Prime number generator

"Title description"
Tmall is designing a cipher system. First he wants to generate some prime numbers. Your task is to help him generate all prime numbers within a given range.
"Input Data"
There are multiple sets of data.
The first line of input data is an integer T (t<=10) that represents the number of groups of test data.
Next T line, each line has two integers m,n, which indicates that the range that is required to generate prime numbers is [M,n] (1<=m<=n<=10^9,n-m<=1000000)
"Output Data"
For each set of test data, output all the prime numbers in [M,n] p, one line at a.
Separate test data with a blank line.
"Sample Input"
2
1 10
3 5
"Sample Output"
2

3

5

7

3

5
"Data Range"
For 30% of data,m<n<=1000;
For 50% of data, m<n<=1000000 and n-m<=1000;
For 100% of data, m<n<=1000000000 and n-m<=1000000;

"Solving" " linear sieve + promiscuity"

"because it's a group of data, and it's going to 10^9, the linear sieve is definitely too much." "

"Since the composite of the 10^9 can be sifted with the sqrt (n) factor, the pretreatment 10^ (4.5) of the mass tables, the linear sieve"

"in M, N in the 10^4.5, you can directly output the prime number within the quality tables, using dichotomy to find the first position of the >=m, start the output can be"

"When M, N is not in the 10^4.5, using the quality tables to sift out composite, marking, because of n-m<=1000000, so the time can be saved into the vis[i-m] so that you can use the array to save. "

#include <cmath> #include <cstdio> #include <cstring> #include <algorithm> #define N 40000 using
namespace Std;
int prime[40010],n,m,t;
BOOL p[40010],vis[1000010];
	 	inline void Shai () {for (int i=2;i<=n;++i) {if (!p[i]) prime[++prime[0]]=i;
	 	 	for (int j=1;j<=prime[0];++j) {if (i*prime[j]>n) break;
	 	 	P[i*prime[j]]=1; if (! (
		  I%PRIME[J])) break;
	}}}-inline int find (int l,int r,int x) {int mid,sum=0;
	 	while (l<=r) {mid= (l+r) >>1;
	 	 if (prime[mid]>=x) sum=mid,r=mid-1;
	 else l=mid+1;
} return sum;
	} int main () {freopen ("prime.in", "R", stdin);
	Freopen ("Prime.out", "w", stdout);
	int i,j;
	Shai ();
	scanf ("%d", &t);
	 	while (t--) {scanf ("%d%d", &m,&n);
			if (n<=n) {int l;
	 	 	L=find (1,PRIME[0],M);
	 	 	  for (int i=l;i<=prime[0];++i) if (prime[i]>n) break;
	 	 	else printf ("%d\n", Prime[i]);
	 	 	printf ("\ n");
		  Continue
		} int l=m+ (m&1);
		if (l==2) l+=2; for (i=l;i<=n;i+=2) Vis[i-m]=1;
		int d=sqrt (n) +1;
		  for (I=2;i<=prime[0];++i) if (prime[i]>d) break;
		 	else {j= (m-1)/prime[i]+1; if (! (
		 	j&1)) J + +;
		 	if (j==1) j+=2;
		 	int r=n/prime[i];
		 for (int k=j;k<=r;++k) vis[k*prime[i]-m]=1;
		} for (int i=m;i<=n;++i) if (!vis[i-m]) printf ("%d\n", I);
		memset (vis,0,sizeof (VIS));
	 printf ("\ n"); }
}