Classic application of POJ 2638 Embedding Method

I won't talk about this question for a long time. I first got a wrong question and thought that the maximum value range is 0 ~ 1000000; I handed in WA and read the question carefully. It turns out that I am so naive... the data is so big,
We found that it only needs to take the continuous 1000000 range value, but we know that the lower part of the 2,147,483,647 open, if it cannot be 0 ~ The division of the quality factor in 2,147,483,647 ^ (1/2), it must be
Prime Number ~ All the prime numbers in the 2,147,483,647 ^ (1/2) interval are filtered out, and then each prime factor is traversed ~ All values in the U range are deleted by multiples of the prime factor p.
Dis [] is marked, and the final discretization is also done... wa I have returned multiple times to note that 1 is not a prime number...
Code:
# Include <iostream>
# Include <cstdio>
Using namespace std;
# Define manx 500000
Bool s [manx];
Int p [manx], num, dis [manx * 2 + 50];
Void prime (){
For (int I = 0; I <manx; I ++)
S [I] = 0;
For (int I = 2; I * I <manx; I ++) {// filter prime numbers
If (! S [I]) {
For (int j = 2; j * I <manx; j ++)
S [I * j] = 1;
}
}
Num = 0;
For (int I = 2; I <manx; I ++)
If (! S [I]) p [num ++] = I;
}
Int main (){
Long a, B;
Prime ();
While (cin> a> B ){
For (int I = 0; I <B-a + 3; I ++)
Dis [I] = 0;
Long min = 1000001, max =-1, l1 = 0, r1 = 0, l2 = 0, r2 = 0;
For (int I = 0; I <num; I ++) {// Delete is a multiple of p [], that is, mark it
Long j = a/p [I];
If (j * p [I]> B) break;
If (a % p [I]) j ++;
If (j = 1) j ++; // a must be a prime number if only 1 can be obtained by division.
For (; j * p [I] <= B; j ++) {// delete a multiple of p [I]
Dis [j * p [I]-a] = 1;
}
}
Long g = 0, left;
For (long I = 0; I <= B-a; I ++ ){
If (! G &&! Dis [I]) {
Left = I;
If (left + a = 1) continue; // 1 is not a prime number. You must note that WA has returned many times.
G = 1; continue;
}
If (! Dis [I]) {
If (min> I-left ){
Min = I-left;
L1 = left; r1 = I;
}
If (max <I-left ){
Max = I-left;
L2 = left; r2 = I;
}
Left = I;
}
} Www.2cto.com
If (min = 1000001 & max =-1) printf ("There are no adjacent primes. \ n ");
Else printf ("% lld, % lld are closest, % lld, % lld are most distant. \ n ", l1 + a, r1 + a, l2 + a, r2 + );
}
}