HDU 2866 - Special Prime（推导）

Special Prime Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 680    Accepted Submission(s): 360   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 Hint   Source 2009 Multi-University Training Contest 7 - Host by FZU

还是那句话：也不是在下不谦虚，要是不看题解，我这辈子都写不出来。

思路：题解源于acdreamers

题意：在区间[2,L]内，有多少个素数p，满足方程有解。

分析：n^b + p*n^(b-1) = m^b   ==>   n^(b-1)*[n+p]=m^b

因为n里面要么有p因子，要么没有，所以gcd(n^(b-1),n+p)=1或(含有p因子的数)

当gcd(n^(b-1),n+p)== (含有p因子的数)的时候，显然无解，因为假设有解，那么n=K*p , K^(b-1)*p^b*(K+1)

如果希望上面的==m^b,那么K^(b-1) *(K+1)必须能表示成某个数X的b次方，而gcd(K,K+1)=1,所以他们根本就没共同因

子，所以没办法表示成X的b次方，所以gcd(n^(b-1),n+p)=1

假设n=x^b，n+p=y^b，那么显然m=x^(b-1)*y，而p=y^b-x^b

显然(y-x)|p，那么必须有y-x=1，所以y=x+1，代上去就发现，p=(x+1)^b-x ^b。所以枚举x,然后判断p是否是素数即可。

这里就可以推出来 p = 3*x^2 + 3*x + 1;

#include <bits/stdc++.h>
using namespace std;
#define clr(a) memset(a,0,sizeof(a))
#define line cout<<"---------------------"<<endl;

typedef long long ll;
const int maxn = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const int Mod = 1e9 + 7;
const int N = 1010;

bool prime[maxn];
void Is_prime(){
	memset(prime,true,sizeof(prime));
	prime[0] = prime[1] = false;
	for(int i=2;i*i<=maxn;i++){
		if(prime[i]){
			for(int j=2;i*j<=maxn;j++){
				prime[i*j] = false;
			}
		}
	}
}
int main(){
    int n;
    Is_prime();
    while(scanf("%d",&n)!=EOF){
        int num=0;
        for(int i=1;;i++){
            int tmp = 3*i*i + 3*i + 1;
            if(tmp > n) break;
            if(prime[tmp])
             	num++;
        }
        if(!num) printf("No Special Prime!\n");
        else printf("%d\n",num);
    }
    return 0;
}