圆上的整点

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int res;
ll gcd(ll a, ll b)
{
	return b == 0 ? a : gcd(b, a % b);
}
bool check(ll n)
{
	ll sq = sqrt(n);
	return sq * sq == n;
}
void solve(ll rr)
{
	for (ll i = 1;i*i<= rr; i++)
	//找i*i+j*j=rr,其中i<j ,并且i*i与j*j互质 
	{
		ll t = rr - i * i;
		if (!check(t))
		//t必须是完全平方数 
		     continue;
		ll j = sqrt(rr - i * i);
		if (i >= j)
		     break;
		if (gcd(i*i, t) == 1)
		 
			res++;
	}
}
int main()
{
	ll r;
	scanf("%lld", &r);
	res = 1;
	r <<= 1;
	for (ll d = 1;d*d <= r; d++)
	//枚举d 
	{
		if (r% d != 0) 
		     continue;
		solve(r / d); //a*a+b*b=r/d
		if (d*d== r) break;
		solve(d);//a*a+b*b=d
	}
	printf("%dn", res << 2);
	return 0;
}

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
int n,ans=1;
int fpow(int a,int b,int MOD)
{
    int s=1;
    while(b){if(b&1)s=1ll*s*a%MOD;a=1ll*a*a%MOD;b>>=1;}
    return s;
}
bool Miller_Rabin(int n)
{
    if(n==2)return true;
    for(int tim=10;tim;--tim)
    {
        int a=rand()%(n-2)+2,p=n-1;
        if(fpow(a,p,n)!=1)return false;
        while(!(p&1))
        {
            p>>=1;int nw=fpow(a,p,n);
            if(1ll*nw*nw%n==1&&nw!=1&&nw!=n-1)return false;
        }
    }
    return true;
}
vector<int> fac;
int Pollard_rho(int n,int c)
{
    int i=0,k=2,x=rand()%(n-1)+1,y=x;
    while(233)
    {
        ++i;x=(1ll*x*x%n+c)%n;
        int d=__gcd((y-x+n)%n,n);
        if(d!=1&&d!=n)return d;
        if(x==y)return n;
        if(i==k)y=x,k<<=1;
    }
}
void Fact(int n,int c)
{
    if(n==1)return;
    if(Miller_Rabin(n)){fac.push_back(n);return;}
    int p=n;while(p>=n)p=Pollard_rho(p,c--);
    Fact(p,c);Fact(n/p,c);
}
int main()
{
    cin>>n;Fact(n,233);sort(fac.begin(),fac.end());
    for(int i=0,l=fac.size(),pos;i<l;i=pos+1)
    {
        int cnt=1;
        pos=i;while(pos<l-1&&fac[i]==fac[pos+1])++pos,++cnt;
        if(fac[i]==2)continue;
        if(fac[i]%4==1)ans=ans*(cnt*2+1);
    }
    printf("%dn",ans*4);
    return 0;
}