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我的更新时间主要在每周4、（5）、6.
且更新内容会偏向英文题目
以CF和UVA为主
同时，感谢洛谷提供的翻译

📣Non-square Equation

🔥题目描述
❄️Let’s consider equation:

❄️ x 2 + s ( x ) ⋅ x − n = 0 x^2+s(x)⋅x−n=0 x2+s(x)⋅x−n=0, where $x,n$ are positive integers, $s(x)$ is the function, equal to the sum of digits of number $x$ in the decimal number system.

❄️You are given an integer $n$ , find the smallest positive integer root of equation $x$ , or else determine that there are no such roots.
🔥输入格式
❄️A single line contains integer n ( 1 < = n < = 1 0 18 ) n (1<=n<=10^{18}) n(1<=n<=1018) — the equation parameter.

❄️Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
🔥输出格式
❄️Print -1, if the equation doesn’t have integer positive roots. Otherwise print such smallest integer $x(x>0)$ , that the equation given in the statement holds.

🔥说明/提示
❄️In the first test case $x=1$ is the minimum root. As $s(1)=1$ and 1 2 + 1 ⋅ 1 − 2 = 0 1^{2}+1·1-2=0 12+1⋅1−2=0

❄️In the second test case $x=10$ is the minimum root. As $s(10)=1+0=1$ and 1 0 2 + 1 ⋅ 10 − 110 = 0 10^{2}+1·10-110=0 102+1⋅10−110=0.

❄️In the third test case the equation has no roots.
🔥说明翻译
❄️在第一组数据中， $x=1$ 是最小根。因为$s(1)=1$ and 1 2 + 1 ⋅ 1 − 2 = 0 1^{2}+1·1-2=0 12+1⋅1−2=0

❄️在第二组数据中，$x=10$是最小根。因为$s(10)=1+0=1$ and 1 0 2 + 1 ⋅ 10 − 110 = 0 10^{2}+1·10-110=0 102+1⋅10−110=0.

❄️在第三组数据中，方程无根。

🔥输入输出样例

❄️输入 #12❄️输出 #11❄️输入 #2110❄️输出 #210❄️输入 #34❄️输出 #3-1

💯AC CODE

#include#define int long long //与signed main()连用 using namespace std;int n;int s(int x){int sum=0;while(x){sum+=x%10;x/=10;}return sum; }signed main()//防止报错 {cin>>n;for(int i=max((double)sqrt(n)-81,(double)1);i<=sqrt(n);++i){if((i*i)+s(i)*i==n){ cout<<i;return 0; }}cout<<"-1"; return 0;}

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