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A Cubic number and A Cubic Number
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
Sample Output
NO NO NO YES NO NO NO YES NO NO
题意:给你一个质素,问这个质数是否可以等于两个立方和数的差,是输出yes,否则输出no。
心路历程:我们不妨设质数为p=x^3-y^3,由立方和公式得p=(x-y)(x^2+xy+y^2),由于p是质数,所以我们可得x-y等于1,
x^2+xy+y^2=p。两式联立得p=3x^2+3x+1=3x(x+1)+1。观察到p-1 是3得倍数,且(p-1)/3可以由两个连续的数的乘积,所以我们只需要判断p-1是否满足这两个条件即可(或者由p=3x^2+3x+1知x正半轴是单调的,可以用二分查找整数根,参考
大佬)
ac代码如下:
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
int ca;
scanf("%d",&ca);
while(ca--)
{
long long m;
scanf("%lld",&m);
if((m-1)%3!=0)//先判断m-1是否是3的倍数
{
printf("NO\n");
continue;
}
long long tmp=(m-1)/3;
long long s=sqrt(tmp);//可以证明若它可以被两个相邻整数的乘积所表示,那么它的平方根的整数部分一定是相邻数中小的那个
if(s*(s+1)==tmp)printf("YES\n");
else
printf("NO\n");
}
return 0;
}