【codeforces】Non-square Equation（数学推导）

最新推荐文章于 2021-01-07 17:21:00 发布

better_space

最新推荐文章于 2021-01-07 17:21:00 发布

阅读量475

点赞数

CC 4.0 BY-SA版权

分类专栏：趣味数学问题 Codeforces 我的收藏文章标签：数学

本文链接：https://blog.youkuaiyun.com/better_space/article/details/53088851

我的收藏同时被 3 个专栏收录

38 篇文章

订阅专栏

Codeforces

33 篇文章

订阅专栏

趣味数学问题

13 篇文章

订阅专栏

本文介绍了一种解决特定形式方程x² + s(x) * x - n = 0的方法，其中s(x)为x的各位数字之和。通过预先估计x的范围，并遍历可能的s(x)值来寻找最小正整数解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Let's consider equation:

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.

Input

A single line contains integer 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.

Output

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.

Sample test(s)

input
2

output
1

input
110

output
10

input
4

output
-1

Note

In the first test case x = 1 is the minimum root. As s(1) = 1 and 12 + 1·1 - 2 = 0.

In the second test case x = 10 is the minimum root. As s(10) = 1 + 0 = 1 and 102 + 1·10 - 110 = 0.

In the third test case the equation has no roots.

本题用二分是不行的，因为y=x+s(x)*x并不是单调递增的。

我们看直接暴力肯定会超时的。所以要看怎么优化，这就需要用到一些数学知识了。看下面这组公式：

我们估算x不会超过10^9,所以s(x)不会超过81；

由上述公式我们就可以对s(x)的值遍历一遍，求出相应的x,再检验此x值是否符合所给公式。从而避免了直接对x遍历超时的情况

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#define MAXN 10000000000
using namespace std;
typedef long long LL;
LL solve(LL x){
	LL t=0;
	while(x){
		t=t+x%10;
		x=x/10;
	}
	return t;
}
int main(){
	LL n;
	while(scanf("%lld",&n)!=EOF){
		LL ans=MAXN;
		for(int i=1;i<=81;i++){
			LL x=sqrt(i*i/4+n)-i/2;
			LL s=solve(x);
			if(x*x+x*s-n==0){
				ans=min(ans,x);
				break;
			}
		}
		if(ans!=MAXN)
			printf("%lld\n",ans);
		else
			printf("-1\n");
	}
	return 0;
}