Hoj question dual screening method

Question:
Returns an interval of <= 1000 000; returns the smallest and largest difference between the adjacent prime numbers of prime numbers;
Analysis:
Double screening, where the number of intervals may be large, so it is obviously not feasible to use Trial Division;
First, the prime number from 1 to 47000 is filtered out for the first time.
Determine whether the number in the range from L to U is a prime number;
Because the range is 1 <= L <u <= 2,147,483,647, while the U-L <= 1000 000; so we can set a 1000000
Array within. The res [I-l] value is 0 or 1 to indicate whether it is a prime number;
CoreAlgorithmIs the second screening, where J = begin * prime [I] indicates the prime [I] in the range A and B
The first multiple; filtering in turn;
Heavy screening methodProgramAs follows:
Memset (Prime, false, sizeof (PRIME ));
TOT = 0;
For (INT I = 2; I <X; I ++)
{
If (! Prime [I])
Isprime [tot ++] = I;
For (Int J = I + I; j <X; j = J + I)
Prime [J] = true;
}

The procedure of double screening is as follows:
Memset (Res, false, sizeof (RES ));
If (M = 1)
M = 2;
For (INT I = 0; I <tot & isprime [I] * isprime [I] <= N; I ++)
{
Long long temp = (M/isprime [I]) + (M % isprime [I]> 0 );
If (temp = 1)
Temp = 2;
For (long J = temp * isprime [I]; j <= N; j + = isprime [I])
Res [J-M] = true;
}

# include
# include
# include
using namespace STD;
# define x 50000
int tot;
bool prime [50000];
int isprime [10000];
bool res [1000010];
void make_prime ()
{< br> memset (Prime, false, sizeof (PRIME);
tot = 0;
for (INT I = 2; I {< br> If (! Prime [I])
isprime [tot ++] = I;
for (Int J = I + I; j prime [J] = true;
}< BR >}< br> int main ()
{< br> freopen ("sum. in "," r ", stdin);
freopen (" sum. out "," W ", stdout);
make_prime ();
long n, m;
while (CIN> m> N)
{< br> memset (Res, false, sizeof (RES);
int CNT = 0;
If (M = 1)
m = 2;
for (INT I = 0; I {< br> long temp = (M/isprime [I]) + (M % is Prime [I]> 0);
If (temp = 1)
temp = 2;
for (long J = temp * isprime [I]; j <= N; j + = isprime [I])
res [J-M] = true;
}< br> for (INT I = 0; I <= N-m; I ++)
If (! Res [I])
CNT ++;
cout }< br> return 0;
}