【codeforces】Non-square Equation(数学推导)

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time limit per test
 1 second
memory limit per test
 256 megabytes
input
 standard input
output
 standard output

Let's consider equation:

x2 + s(xx - 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 cincout 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 10000000000using 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);elseprintf("-1\n");}return 0;} 




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