Linear appreciated screen

Linear sieve

Idea: Each number has one and only one screen
Resolution: Each number is only (non-native) screen factor to its maximum

Together provided the smallest prime factor of the number of x p, x = pq, is easy to know p <= q

We let q x is only weed out, choose an enumeration q

Enumeration q, retrieve all the p, weed out the number of x, left out

void sieve() {
	for(int i=2;i<=m;i++) { //枚举q 
	    if(v[i]==0) {
	    	ps[++cnt]=i; //统计p 
		}
		for(int j=1;j<=cnt;j++) { //枚举满足的p 
			if(i*ps[j]>m) break;
			v[i*ps[j]]=1;
			if(i%ps[j]==0) break; //条件判断 
		}
    }
} //统计2到m的质数 

The code means: Let q be the smallest prime factor of u, q = uv

1. If p <= u (p is a prime number), then p is the smallest prime factor of x = pq and satisfies the condition, the screen to x

2. If p> u, x = pq = puv = u * (pv), p x is not the smallest prime factor, this number will be PV (greater than q) filtering out, repeat, can not jump out of the screen x

So far, the accuracy and adequacy of the algorithm are proved.

(Original)