思路:
1.p=x^3-y^3
2.p=(x-y)(x^2+xy+y^2)
3.p是素数,因子只有1和自身,而(x^2+xy+y^2)>1
4.(x-y)==1 , x=y+1
5. p=3*y*y+3*y+1
6.计算i~1e6 之间 A[i]=3*i*i+3*i+1
7.二分查找
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
const int N=1e6+5;
typedef long long ll;
ll x[N];
void solve()
{
for(ll i=0;i<N;i+=1)
{
x[i]=3*i*i+3*i+1;
}
}
bool Binary(ll p)
{
int L=1;
int R=N-1;
while(L<=R)
{
int M=(L+R)>>1;
if(x[M]==p)
return true;
else if(x[M]<p)
L=M+1;
else
R=M-1;
}
return false;
}
int main()
{
int T;
scanf("%d",&T);
solve();
while(T--)
{
ll p;
scanf("%lld",&p);
if(Binary(p))
printf("YES\n");
else
printf("NO\n");
}
return 0;
}