Special Prime（数论+思维）

Special Prime

 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

Hint

题意：

给你一个数L，问不超过L的范围内有多少素数可以满足上面是式子，其中mn都是非零自然数

分析：

我们已知式子

$n^3 + p \cdot n^2 = m^3$

观察所给的例子我们发现，好像n都是立方数

我们进一步给式子变形得到：

$n^2(p+n) = m^3$

我们观察到n都是立方数，那么只需要使得$p+n$ 也是立方数就能满足式子

即设$n = b^3,p+n = a^3$

所以 $(b^3)^2\cdot a^3 = (b^2)^3 \cdot a^3 = (b^2a)^3 = m^3$

因此我们可以得到：

$p + b^3 = a^3$

$\downarrow$

$p = a^3 - b^3 = (a - b)(a^2+ab+b^2)$

我们得到了素数p的一个乘积的形式

所以$a - b = 1$

即$a = b + 1$

我们带入原式得到：

$p = (b+1)^2 + (b+1)b + b^2 = 3b^2 + 3b + 1$

我们把右边变成完全平方的形式

两边先同乘12

$12p = 36b^2+36b+12$

$12p = 36b^2+36b+9+3 = (6b+3)^2+3$

所以

$12p-3 = (6b+3)^2$

因此我们只需判断12p-3是否是完全平方数即可其中p是素数

code：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6+10;
bool isprime[maxn];
int prime[maxn];
int cnt;
void init(){
    memset(isprime,true,sizeof(isprime));
    isprime[0] = isprime[1] = false;
    cnt = 0;
    prime[1] = 0;
    for(int i = 2; i < maxn; i++){
        if(isprime[i]){
            int p = 12 * i - 3;
            int tmp = sqrt(p);
            if(tmp * tmp == p) cnt++;
            for(int j = i + i; j < maxn; j += i){
                isprime[j] = false;
            }
        }
        prime[i] = cnt;
    }
}
int main(){
    init();
    int l;
    while(~scanf("%d",&l)){
        if(prime[l] == 0)
            printf("No Special Prime!\n");
        else
            printf("%d\n",prime[l]);
    }
    return 0;
}