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-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\n"); } else{ if(n%4!=0){ printf("0 0\n"); } 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\n",cnt,ans); } } } return 0; }
这题卡long long的。。
代码：
原文：https://www.cnblogs.com/wrjlinkkkkkk/p/10644502.html