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).
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
2
3
5
7
11
13
17
19
23
29
Sample Output
NO
NO
NO
YES
NO
NO
NO
YES
NO
NO
NO
NO
YES
NO
NO
NO
YES
NO
NO
Source
a^3-b^3 == p，p为质数，所以a-b=1
//2017-09-17
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long using namespace std; const int N = ; ll cubic[N+];
ll diff[N+]; bool check(ll p){
int pos = lower_bound(diff+, diff+N, p) - diff;
if(diff[pos] == p)
return true;
return false;
} int main()
{
int T;
scanf("%d", &T);
ll p;
for(ll i = ; i <= N; i++)
cubic[i] = i*i*i;
for(int i = ; i <= N; i++)
diff[i] = cubic[i+]-cubic[i];
while(T--){
scanf("%lld", &p);
if(check(p))
printf("YES\n");
else printf("NO\n");
} return ;
}