hdu 6216 A Cubic number and A Cubic Number

最新推荐文章于 2020-03-18 17:28:54 发布

Ambition_Bit

最新推荐文章于 2020-03-18 17:28:54 发布

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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

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;
}