【Codeforces-233B】-Non-square Equation(思维,公式转换)

    xiaoxiao2021-12-12  2

    B. Non-square Equation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

    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 cincout streams or the %I64dspecifier.

    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.

    Examples 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.

    题解:根据一元二次方程的求根公式可知:x=sqrt(sx^2/4+n)-sx/2;

    而n取最大值时,x<10e9,所以sx<9*9=81;所以只需要枚举sx的值从1到81根据公式求的x的值,再带入判断即可。

    #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> using namespace std; #define LL long long int get(LL x) //求 x 每一位的和 { int sum=0; while(x) { sum+=x; x/=10; } return sum; } int main() { LL n; scanf("%lld",&n); LL ans=-1; for(int i=1;i<=81;i++) // i 表示sx可能值,最大 81 { LL x=sqrt(i*i/4+n)-i/2; //求 sx 对应的 x int sx=get(x); //求出 x 各位的和 if(x*x+sx*x-n==0) { ans=x; break; } } printf("%lld\n",ans); return 0; }

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