Codeforces Round #144 (Div. 2)——B

最新推荐文章于 2025-08-11 09:32:41 发布

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最新推荐文章于 2025-08-11 09:32:41 发布

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本文探讨了一种特殊的非平方方程 x^2 + s(x) * x - n = 0 的最小正整数解问题，其中s(x)为x的各位数字之和。文章提供了一个高效的算法来寻找该方程的最小正整数解，并通过样例验证了算法的有效性。

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 usecin, 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.

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;

int getsum(long long x)//求数字各位之和，由于x(x+s(x))=n,故x最大为10^9，故s(x)最大为81
{
    int s=0;
    while(x)
    {
        s+=x%10;
        x/=10;
    }
    return s;
}
int main()
{
    long long n;
    scanf("%I64d",&n);
    //因为x(x+s(x))=n，sqrt(n)-82<x<sqrt(n)
    long long x=(long long)sqrt(n);
    long long ave=(x-82>0?x-82:1);
    long long ans=-1;
    for(; ave<=x; ave++)
        if(ave*ave+ave*getsum(ave)==n)
            {
                ans=ave;
                break;
            }
    printf("%I64d\n",ans);
    return 0;
}