HDU - 2866- Special Prime【数论】题解-蒲公英云

HDU - 2866- Special Prime【数论】题解

傷城~ 2021-07-25 00:10 467阅读 0赞

1.题目

Give you a prime number p, if you could find some natural number (0 is not inclusive) n and m, satisfy the following expression:
在这里插入图片描述
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 + 71^2 = 2^3
8^3 + 198^2 = 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

2.代码

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int maxn=1e6+10;
int prime[maxn],ans[maxn];
void isprime()
{ 
    prime[0]=1;
    prime[1]=1;
    for(int i=2;i<maxn;i++)
    { 
        if(prime[i]==0)
        { 
            for(int j=i+i;j<maxn;j+=i)
             prime[j]=1;
        }
    }
}
bool  judge(int p)//判断12*P-3是否是平方数 
{ 
    int m=12*p-3;
    int k=(int)sqrt(m);
    return  k*k==m;
}
void table()
{ 
   for(int i=1;i<maxn;i++)
   { 
      if(!prime[i]&&judge(i))    
         ans[i]++;
      ans[i]+=ans[i-1];
   }    
} 
int main()
{ 
    isprime();
    table();
    int  l;
     while(cin>>l)
     { 
        if(ans[l])
         cout<<ans[l]<<endl;
       else
         cout<<"No Special Prime!"<<endl;    
    }
    return 0;
}