【2017 ACM/ICPC Asia Regional Qingdao Online 1011】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): 148    Accepted Submission(s): 89

Problem Description

A cubic number is the result of using a whole number in a multiplication three times. For example,3×3×3=27  so27 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 integerT (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 givenp is a difference of two cubic numbers, or 'NO' if not.

 

Sample Input

102357111317192329

 

Sample Output

NONONOYESNONONOYESNONO

 

题目大意：

给出一个p,p是素数（2 <= p <=1e12），判断是否存在正整数a,b使得a*a*a - b*b*b = p

如果存在输出YES,否则输出NO

我们知道立方差公式：a*a*a  - b*b*b = (a-b)(a*a+a*b+b*b)  

我们知道a*a*a - b*b*b = prime(素数)

所以(a-b)(a*a+a*b+b*b)  = prime;

由因为prime 只能由1*prime得来，所以prime = 1*prime

所以（a-b） = 1, (a*a+a*b+b*b) = prime , 所以a,b必是两个相邻的整数

所以预处理+二分即可

#include<bits/stdc++.h>#include <ctime>using namespace std;typedef long long ll;const int MAXN = 1 * 1e5 + 500;const ll M = 1e9 + 7;vector<ll> v;ll sum(ll x){    return x * x * x;}void init(){    ll limt = 1e12 + 1;    for (int i = 2;; i++)    {        ll temp = sum(i) - sum(i - 1);        if (temp > limt)        {            break;        }        v.push_back(temp);    }}int main(){    ///clock_t start_time = clock();    ///clock_t end_time = clock();    ///cout << "Running time is: " << static_cast<double>(end_time - start_time) / CLOCKS_PER_SEC * 1000 << "ms" << endl;    std::ios::sync_with_stdio(false);    int t;    cin >> t;    init();    while (t--)    {        ll p;        cin >> p;        ll flag = v[lower_bound(v.begin(), v.end(), p) - v.begin()];        if (p != flag)        {            cout << "NO" << endl;        }        else        {            cout << "YES" << endl;        }    }    return 0;}