hdu 6216 A Cubic number and A Cubic Number

ゝ一纸荒年。 2022-06-08 07:06 157阅读 0赞

# A Cubic number and A Cubic Number #

**Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)  
Total Submission(s): 438 Accepted Submission(s): 237**

Problem Description

A cubic number is the result of using a whole number in a multiplication three times. For example,  3×3×3=27 so  27 is a cubic number. The first few cubic numbers are  1,8,27,64 and  125. Given an prime number  p. Check that if  p is a difference of two cubic numbers.

Input

The first of input contains an integer  T(1≤T≤100) which is the total number of test cases.  
For each test case, a line contains a prime number  p(2≤p≤1012).

Output

For each test case, output 'YES' if given  p is a difference of two cubic numbers, or 'NO' if not.

Sample Input

10
    2
    3
    5
    7
    11
    13
    17
    19
    23
    29

Sample Output

NO
    NO
    NO
    YES
    NO
    NO
    NO
    YES
    NO
    NO
      
       
      
       
    
      
       
      
       
    
      
       
      
       
    
      
       
      
       
        
       题目大意：
      
       
      
       
        
       题目的意思就是说，给你一个素数P，问是否有两个数的立方差等于p，如果有输出YES，没有输出NO；
      
       
      
       
    
      
       
      
       
    
      
       
      
       
        
       题目解析：
      
       
      
       
        
       由题目可以得到，由于a^3-b^3=(a-b)(a^2+b^2+ab) 所以（a-b）a-b只有等于1，a^3-b^3的差才会是素数，所以可以得到
      
       
      
       
        
       （i+1）^3-i^3;
      
       
      
       
    
      
       
      
       
    
      
       
      
       
        
       代码：
      
       
      
       
       
        #include <cstdio>
    #include <iostream>
    using namespace std;
    int main()
    {
        //freopen("in.txt","r",stdin);
        int T;
        cin>>T;
        while(T--)
        {
           long long n;
            cin>>n;
            for(long long i=1;i<100000000;i++){
               if((i+1)*(i+1)*(i+1)-i*i*i>n){
                    cout<<"NO"<<endl;
                    break;
               }
               if((i+1)*(i+1)*(i+1)-i*i*i==n){
                    cout<<"YES"<<endl;
                    break;
               }
            }
        }
        return 0;
    }