Hit 2276 (number theory, prime number )]

Hit 2276 (number theory, prime number )]

 

[Original question link]

Http://acm-hit.sunner.cn/index.php? Option = com_wrapper & Itemid = 39

(Not accurate. You need to go to hit to find the question number)

 

[Topic]

Number of prime numbers between input L, R output L, and R (L <= r <= 2147483647, R-l <= 1000000)

 

[Solution Thinking]

The data range is relatively large. We need to make full use of the method to determine the prime number. We can know that if a number X is not a prime number, and only if there is a number y less than or equal to SQRT (x, so that X % Y = 0, so SQRT (2147483647) can be used to make the overall judgment, because the R-L has a range, therefore, we can traverse the number sieve in this range to remove a multiple of the prime numbers smaller than SQRT (R). What is not screened is the prime number. Just traverse the number of prime numbers.

 

[Code]

# Include <iostream>

# Include <cmath>

Using namespace STD;

 

Unsigned long prime (unsigned long a [], unsigned long N)

{

Unsigned long I, J, K, X, num, * B;

N ++;

N/= 2;

B = new unsigned long [(n + 1) * 2];

A [0] = 2; A [1] = 3; num = 2;

For (I = 1; I <= 2 * n; I ++)

B [I] = 0;

For (I = 3; I <= N; I + = 3)

For (j = 0; j <2; j ++)

{

X = 2 * (I + J)-1;

While (B [x] = 0)

{

A [num ++] = X;

For (k = x; k <= 2 * n; k + = X)

B [k] = 1;

}

}

Return num;

}

 

Unsigned long PR [50000];

Bool ishave [1100000];

 

Int main ()

{

Unsigned long l, R, I, j, ans;

Prime (PR, 50000 );

While (scanf ("% LD % lD", & L, & R )! = EOF)

{

Ans = 0;

For (I = 0; I <= R-l; I ++) ishave [I] = 1;

For (I = 0; PR [I] * PR [I] <= r; I ++)

{

If (L % PR [I]) J = (L/PR [I] + 1) * PR [I];

Else J = L;

For (; j <= r; j + = Pr [I])

{

If (J! = Pr [I]) ishave [J-L] = 0;

}

}

For (I = 0; I <= R-l; I ++)

If (ishave [I]) ans ++;

Printf ("% LD \ n", ANS );

}

Return 0;

}

[\ Code]