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：

$displaystyle x-sqrt{n}=z+y-2sqrt{yz}$
$if sqrt{n}quad isquad rational quad number:$
$qquad atquad leastegin{Bmatrix}
x=sqrt{n}+z\
y=0
end{Bmatrix}$
$else egin{Bmatrix}
n = 4yz\
x=y+z\
x^2 geq n
end{Bmatrix} quad infty$
$whichquad is$
$displaystyle
egin{Bmatrix}
yz = frac{n}{4}\
(y+z) geq n
end{Bmatrix}$

#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-6;
const int mod = 1e9+7;
const int maxn = 2e6+100;
const int maxm = 2e6+100;
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
");
        }
        else{
            if(n%4!=0){
                printf("0 0
");
            }
            else{
                int cnt = 0;
                ll ans = 0;
                int m = sqrt(n/4+0.5);
                int c = sqrt(m);
                for(int i = 1; i <= m; i++){
                    if((n/4)%i==0&&((i+(n/4)/i)>=c)){

                        ans += 1ll*i*(n/4)/i*(i+(n/4)/i);
                        cnt++;
                        ans%=mod;
                    }
                }
                printf("%d %lld
",cnt,ans);
            }
        }
    }
    return 0;
}

这题卡long long的。。

代码：