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：