Non-square Equation【公式转换】
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Non-square Equation
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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 Input
Input
2
Output
1
Input
110Output
10
Input
4Output
-1
Hint
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.
【题意】根据公式,给出n,求能满足公式的最小正整数,S(x) 为x每一位数字的和, 不存在x满足公式时,输出-1。
【思路】公式转换
s(x) 函数,可以发现 s(x) 在 x 取得最大值,即 x = 999999999 时相应地取得最大值,即 81。于是,我们可以去枚举 s(x) ,计算相应的 x 能否使方程左右两边相等,这样就能大大降低复杂度。利用配方公式,可以得到x = sqrt( S(x)^2 /4 + n ) - S(x)/2。如此一来,我们只需要从 1 到 81 枚举所有 s(x),然后找到对应的最小x,代入方程检查是否相等符合。
AC 代码:
#include<cmath>#include<cstdio>using namespace std;int get(__int64 x) { int sum = 0; while(x) { sum += x%10; x /= 10; } return sum;}int main() { __int64 n, ans = -1; scanf("%I64d", &n); for(int i=1; i<=81; ++i) { __int64 x = sqrt(i*i/4+n) - i/2; int sx = get(x); if(x*x+sx*x-n == 0) { ans = x; break; } } printf("%I64d\n", ans); return 0;}
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