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Special Prime

2024-08-15 09:00:33   218人阅读

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:
  
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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；
 1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<queue> 5 #include<set> 6 #include<math.h> 7 #include<string.h> 8 using namespace std; 9 typedef long long LL;10 bool prime[1000005];11 int sum[1000005];12 int main(void)13 {14         int i,j;15         memset(sum,0,sizeof(sum));16         for(i = 2; i <=1000; i++)17         {18                 if(!prime[i])19                 {20                         for(j = i; (i*j) <= 1000000; j++)21                         {22                                 prime[i*j] = true;23                         }24                 }25         }26         for(i = 0;; i++)27         {28                 int x = 3*i*i+3*i+1;29                 if(x > 1e6)30                         break;31                 if(!prime[x])32                 {33                         sum[x] = 1;34                 }35         }sum[1] = 0;36         for(i = 1; i <= 1e6; i++)37         {38                 sum[i] += sum[i-1];39         }40         int n;41         while(scanf("%d",&n)!=EOF)42         {       if(sum[n])43                 printf("%d\n",sum[n]);44                 else printf("No Special Prime!\n");45         }46         return 0;47 }

 

 

Special Prime


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