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.
Example 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 <cstdio> #include <cmath> #define ll long long int get(int i){ int ans = 0; while (i){ ans +=i; i/=10; } return ans; } int main(){ ll n; scanf("%lld",&n); ll ed = -1; for (ll i = 1; i <= 162; ++i){ //无脑直接让x的18位都是9 这样sx最大也只是162 其实x最大只能取到1e9左右 所以sx取81 ll at = sqrt(i*i+4*n)-i; //直接枚举sx的情况 at/=2; //获得x的值 将x带入公式判断即可 ll sx = get(at); if (at*at+sx*at-n==0){//不能直接为i 因为at可能不整数 而定义的at是直接取整的 如果at不是整数就不成立 所以再对at求sx判断是否成立 ed = at; break; } } printf("%lld\n",ed); return 0; }