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): 910    Accepted Submission(s): 444

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

102357111317192329

 

Sample Output

NONONOYESNONONOYESNONO

 

Source

2017 ACM/ICPC Asia Regional Qingdao Online

 

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liuyiding

题意：给一个素数n，n是否能通过a^3-b^3得到

a^3-b^3=(a-b)*(a^2+a*b+b^2)，要想a^3-b^3=素数，(a-b)=1，所以满足要求的数一定是两个相邻的数的立方差，(i)^3-(i-1)^3=(3i^2-3*i+1)，这个式子不能继续化简为两个式子相乘，所以i^3-(i-1)^3必为素数，可以先打表到两个立方差到1e12，i的范围大概为577352，数据范围较小，直接暴力比较即可

#pragma comment(linker,"/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<stack>#include<queue>#include<deque>#include<set>#include<map>#include<cmath>#include<vector>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair<int, int> PII;#define pi acos(-1.0)#define eps 1e-10#define pf printf#define sf scanf#define lson rt<<1,l,m#define rson rt<<1|1,m+1,r#define e tree[rt]#define _s second#define _f first#define all(x) (x).begin,(x).end#define mem(i,a) memset(i,a,sizeof i)#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)#define mi ((l+r)>>1)#define sqr(x) ((x)*(x))#define sq(x) (3*(x)*(x)-3*(i)+1)const int inf=0x3f3f3f3f;int t;ll n,q[10000000];void dabiao(){    ll i;    for(i=2;i<=577352;i++)    {        q[i]=sq(i);    }}int main(){    dabiao();    sf("%d",&t);    while(t--)    {        int tag=0;        sf("%lld",&n);        for(int i=2;i<=577351;i++)            if(n==q[i])            {                puts("YES");                tag=1;                break;            }        if(!tag)            puts("NO");    }    return 0;}