HDU6216

Posted 2020-10-07 Penn000

A Cubic number and A Cubic Number

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 313    Accepted Submission(s): 184

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

 

a^3-b^3 == p，p为质数，所以a-b=1

 1 //2017-09-17
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <algorithm>
 6 #define ll long long
 7 
 8 using namespace std;
 9 
10 const int N = 600000;
11 
12 ll cubic[N+100];
13 ll diff[N+100];
14 
15 bool check(ll p){
16     int pos = lower_bound(diff+1, diff+N, p) - diff;
17     if(diff[pos] == p)
18       return true;
19     return false;
20 }
21 
22 int main()
23 {
24     int T;
25     scanf("%d", &T);
26     ll p;
27     for(ll i = 1; i <= N; i++)
28         cubic[i] = i*i*i;
29     for(int i = 1; i <= N; i++)
30         diff[i] = cubic[i+1]-cubic[i];
31     while(T--){
32         scanf("%lld", &p);
33         if(check(p))
34           printf("YES\n");
35         else printf("NO\n");
36     }
37 
38     return 0;
39 }