hdu 2866 Special Prime（数论，推公式）

最新推荐文章于 2020-06-16 20:42:45 发布

acm_cxq

最新推荐文章于 2020-06-16 20:42:45 发布

阅读量847

点赞数

分类专栏：数学-数论文章标签： ACM HDU 数学

本文链接：https://blog.youkuaiyun.com/acm_cxq/article/details/51621670

版权

数学-数论专栏收录该内容

39 篇文章

订阅专栏

Special Prime

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 390 Accepted Submission(s): 200

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

Sample Output

思路参考：点击打开链接

这里有个疑问，为什么把n=x³，n+p=y³需要先证明gcd（n²，n+p）=1？

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 1000001
#define LL long long
int prime[N],notprime[N];
void init()
{
    for(int i=2;i*i<N;i++)
    {
        if(!notprime[i])
        {
            for(LL j=(LL)i*i;j<N;j+=i)
                notprime[j]=1;
        }
    }
}
int main()
{
    int n;
    init();
    while(~scanf("%d",&n))
    {
        int num=0;
        for(int i=1;;i++)
        {
            int tmp=3*i*i+3*i+1;
            if(tmp>n) break;
            if(!notprime[tmp])
                num++;
        }
        if(!num) printf("No Special Prime!\n");
        else printf("%d\n",num);
    }
    return 0;
}