【HDU】2866：Special Prime【数论】

Special Prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 738    Accepted Submission(s): 390

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

 

Recommend

gaojie

 

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Solution

纯数论推式子找性质辣

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 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。
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作者：acdreamers
来源：CSDN
原文：https://blog.csdn.net/acdreamers/article/details/8572959 

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 Code

#include<bits/stdc++.h>
using namespace std;

int isnot[1000005], prime[300005], t, ans[1000005];

void init() {
    isnot[1] = 1;
    for(int i = 2; i <= 1000000; i ++) {
        if(!isnot[i])    prime[++t] = i;
        for(int j = 1; j <= t; j ++) {
            int v = prime[j] * i;
            if(v > 1000000)    break;
            isnot[v] = 1;
            if(i % prime[j] == 0)    break;
        }
    }
    for(int i = 1; ; i ++) {
        int v = (i + 1) * (i + 1) * (i + 1) - i * i * i;
        if(v > 1000000)    break;
        if(!isnot[v]) {
            ans[v] = 1;
        }
    }
    for(int i = 1; i <= 1000000; i ++)    ans[i] += ans[i - 1];
}

int main() {
    int n;
    init();
    while(scanf("%d", &n) == 1) {
        if(n < 7)    printf("No Special Prime!
");
        else        printf("%d
", ans[n]);
    }
    return 0;
}