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): 224 Accepted Submission(s): 129

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
题意；给你一个素数p，问它能否由两个立方数的差得到。
题解：设两个数a,b(a>b),则a^3-b^3=(a-b)*(a^2+a*b+b^2).
如果是素数则b=a-1. 原式等于 3*a^2-3*a+1==p,a>=sqrt((p-1)/3)。往后枚举几个数就行了。
代码：

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<vector>#include<queue>#include<stack>#include<set>#include<algorithm>#include<map>#include<time.h>#include<math.h>//#define pb push_back//#define mp make_pair#define INF 0x3f3f3f3fusing namespace std;typedef long long int ll  ;typedef pair<int,int>pp;const int N=1e5+100;const ll mod=1e9+7;int read(){    int x=0;    char ch = getchar();    while('0'>ch||ch>'9')ch=getchar();    while('0'<=ch&&ch<='9')    {        x=(x<<3)+(x<<1)+ch-'0';        ch=getchar();    }    return x;}/***********************************************************/ll p;int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%lld",&p);        bool flag=false;        ll x=sqrt((p-1)/3);        for(ll i=x;i<=x+2; i++)        {            if(3*i*i-3*i+1==p)            {                flag=true;                break;            }            if(3*i+i-3*i+1>p) break;        }        if(flag) puts("YES");        else            puts("NO");    }}