HDU 6216 A Cubic number and A Cubic Number【数学思维+枚举/二分】

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

【题意】：询问一个质数p是否可以写成两个立方数的差。

【分析】：

x^3-y^3
=(x^3-x^2*y)+x^2*y-(y^3-x*y^2)-x*y^2
=x^2(x-y)-y^2(y-x)+xy(x-y)

=(x-y)(x^2+xy+y^2)=p（p是质数）——> x-y=1以及x^2+xy+y^2=p

代入消元：p=3x^2+3x+1

【代码】：

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<map>
#include<set>
#include<string>
using namespace std;

int main()
{
    int t,p,flag;
    scanf("%d",&t);
    while(t--)
    {
        flag=0;
        scanf("%d",&p);
        for(int i=1;i<=1e6+10;i++)
        {
            if(3*i*i+3*i+1==p)
            {
                flag=1;
                break;
            }
        }
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string.h>
#include <string>
#define ll long long  

using namespace std;  

const int MAXN = 1e6 + 5;
ll tab[MAXN];  

void init()
{
    for (ll i = 0;i < MAXN;++i)
        tab[i] = 3 * i*i + 3 * i + 1;
}  

int main()
{
    int T;
    scanf("%d", &T);
    init();
    for (int i = 1;i <= T;++i)
    {
        bool flag = false;
        ll v;
        scanf("%I64d", &v);
        int left = 0, right = MAXN-1;
        int mid = (left + right) >> 1;
        while (left <= right)
        {
            if (v == tab[mid])
            {
                flag = true;
                break;
            }
            else if (v > tab[mid])
                left = mid + 1;
            else
                right = mid - 1;
            mid = (left + right) >> 1;
        }
        if (flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    //system("pause");
    return 0;
} 

预处理+二分查找//参考