Comet OJ Contest #0 解方程(暴力)

题意：

给定自然数n，求满足$\displaystyle \sqrt{x-\sqrt{n}}=\sqrt{z}-\sqrt{y}$的x,y,z，输出解的个数以及所有解 xyz的和

n<=1e9,t<=5000,1500ms

思路：

$\displaystyle x-\sqrt{n}=z+y-2\sqrt{yz}$
$if \sqrt{n}\quad is\quad rational \quad number:$
$\qquad at\quad least\begin{Bmatrix}
x=\sqrt{n}+z\\
y=0
\end{Bmatrix}$
$else \begin{Bmatrix}
n = 4yz\\
x=y+z\\
x^2 \geq n
\end{Bmatrix} \quad infty$
$which\quad is$
$\displaystyle
\begin{Bmatrix}
yz = \frac{n}{4}\\
(y+z) \geq n
\end{Bmatrix}$

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 2e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int main() {
    int t;
    scanf("%d", &t);
    while(t--){
        int n;
        scanf("%d", &n);
        if((int)sqrt(n)*sqrt(n)==n){
            printf("infty\n");
        }
        else{
            if(n%!=){
                printf("0 0\n");
            }
            else{
                int cnt = ;
                ll ans = ;
                int m = sqrt(n/+0.5);
                int c = sqrt(m);
                for(int i = ; i <= m; i++){
                    if((n/)%i==&&((i+(n/)/i)>=c)){

                        ans += 1ll*i*(n/)/i*(i+(n/)/i);
                        cnt++;
                        ans%=mod;
                    }
                }
                printf("%d %lld\n",cnt,ans);
            }
        }
    }
    return ;
}

这题卡long long的。。

代码：