https://www.cnblogs.com/Gloid/p/9538413.html 基本思路没有太大差别。得到2n=d(a2+3b2)，其中d=gcd(n-x,n+x)，n-x==a2&&n+x==3b2||n-x==3a2&&n+x==b2。于是枚举d，然后枚举b。复杂度玄学。
#include#include #include #include #include #include using namespace std;#define ll long longint read(){ int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') { if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f;}char getc(){ char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')) c=getchar();return c;}ll gcd(ll n,ll m){ return m==0?n:gcd(m,n%m);}int T;ll n,m;bool issqr(ll n){ return (ll)(sqrt(n))*(ll)(sqrt(n))==n;}int calc(ll n,ll d){ int s=0; for (int i=1;d*i*i<=n;i++) { ll b=1ll*i*i,a=m/d-b; if (a%3==0&&issqr(a/3)&&gcd(a,b)==1) s++; } for (int i=1;3*d*i*i<=n;i++) { ll b=3ll*i*i,a=m/d-b; if (issqr(a)&&gcd(a,b)==1) s++; } return s;}int main(){#ifndef ONLINE_JUDGE freopen("bzoj4544.in","r",stdin); freopen("bzoj4544.out","w",stdout); const char LL[]="%I64d\n";#else const char LL[]="%lld\n";#endif T=read(); while (T--) { cin>>n;m=n<<1; int ans=0; for (int d=1;1ll*d*d<=m;d++) if (m%d==0) { ans+=calc(n,d); if (m/d!=d) ans+=calc(n,m/d); } ans*=4;ans+=2; cout< <