HDU—6216 A Cubic number and A Cubic Number

A Cubic number and A Cubic Number
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2572 Accepted Submission(s): 1050

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

Source
2017 ACM/ICPC Asia Regional Qingdao Online

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【分析】
题意：给出一个素数p，问p是否是两个立方数的差。
那么设存在整数a,b使得 $a^3 - b^3 = p$
化简得$(a-b)*(a^2 + ab + b^2) = p$
因为p是素数，所以p应该只有两个因子1和p本身，那么显然$a^2 + ab + b^2 = p$ , $a-b = 1$
所以可以证明a,b相差1，即$a = b + 1$，代入得到$3b^2 + 3b + 1 = p$
然后就是解方程的事情了，只要能解出一个整数解就表示存在两个立方数之差等于p
【代码】

#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

int main (){
  int pp;scanf("%d",&pp);
  while (pp--){
    long long n;scanf("%lld",&n);
    long long delta = 9 + 4 * 3 * (n - 1);
    long long sq = sqrt(delta);
    if (sq * sq == delta){
      long long x1 = -sq - 3;
      if (x1 % 6 == 0){
        puts("YES");
        goto out;
      }
      long long x2 = sq - 3;
      if (x2 % 6 == 0){
        puts("YES");
        goto out;
      }
    }
    puts("NO");
    out:;
  }
  return 0; 
}