Special Prime

Special Prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 415    Accepted Submission(s): 220

Problem Description

Give you a prime number p, if you could find some natural number (0 is not inclusive) n and m, satisfy the following expression:
  
We call this p a “Special Prime”.
AekdyCoin want you to tell him the number of the “Special Prime” that no larger than L.
For example:
  If L =20
1^3 + 7*1^2 = 2^3
8^3 + 19*8^2 = 12^3
That is to say the prime number 7, 19 are two “Special Primes”.

 

Input

The input consists of several test cases.
Every case has only one integer indicating L.(1<=L<=10^6)

 

Output

For each case, you should output a single line indicate the number of “Special Prime” that no larger than L. If you can’t find such “Special Prime”, just output “No Special Prime!”

 

Sample Input

7

777

Sample Output

1

10

思路：假设 n^2 和 n+p 之间有公共素因子 p , 那么 n+p = k*p , 即 n=p*（k-1）,带进去得到 p^3 * (k-1)^2 *k = m^3 , (k-1)^2*k 肯定是不能表示成某一个数的三次幂的，所以假设不成立,gcd(K,K+1)=1,假设n=x^3 , n+p=y^3 , 相减得到 p = y^3 - x^3 = (y-x) *(y^2+y*x+x^2)p是素数，y-x=1 ， p =(x+1)^3 - x^3 = 3*x^2+3*x+1,然后枚举x，判断求得数是否是素数即可；

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<set>
#include<math.h>
#include<string.h>
using namespace std;
typedef long long LL;
bool prime[1000005];
int sum[1000005];
int main(void)
{
        int i,j;
        memset(sum,0,sizeof(sum));
        for(i = 2; i <=1000; i++)
        {
                if(!prime[i])
                {
                        for(j = i; (i*j) <= 1000000; j++)
                        {
                                prime[i*j] = true;
                        }
                }
        }
        for(i = 0;; i++)
        {
                int x = 3*i*i+3*i+1;
                if(x > 1e6)
                        break;
                if(!prime[x])
                {
                        sum[x] = 1;
                }
        }sum[1] = 0;
        for(i = 1; i <= 1e6; i++)
        {
                sum[i] += sum[i-1];
        }
        int n;
        while(scanf("%d",&n)!=EOF)
        {       if(sum[n])
                printf("%d
",sum[n]);
                else printf("No Special Prime!
");
        }
        return 0;
}