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CodeForces 233B Non-square Equation


题目:
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 cin, cout 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.
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.

 

分析与总结:
之前很少做数学题,所以一看到就想用二分法做,但是一直WA在test 5,后来经提醒可以将公式变形,瞬间明朗。

把公式x2 + s(x)·x - n = 0, 进行变形:
S(x) = n/x - x。
可大致估计S(x)的范围在1~100之间, 然后枚举S(x)的值,根据S(x)的值和方程S(x) = n/x - x,解出x = sqrt( S(x)^2/4 + n ).
然后把x代入原公式x2 + s(x)·x - n = 0 看是否符合。


代码:
[cpp] 
#include<iostream> 
#include<cstdio> 
#include<cmath> 
using namespace std; 
 
typedef long long int64; 
int64 n, sx; 
 
int64 digitSum(int64 n){ 
    int64 sum=0; 
    while(n){ 
        sum += n%10; 
        n/=10; 
    } 
    return sum; 

 
int main(){ 
    while(cin >> n){ 
        int64 x=1, end=1e8, ans=-1; 
         
        for(int64 i=1; i<=100; ++i){ 
            int64 tmp = i*i/4+n; 
 
            x = sqrt(tmp)-i/2;  
 
            sx = digitSum(x); 
            if(x*x+sx*x-n==0){ 
                ans=x; 
                break; 
            } 
        } 
        cout << ans << endl; 
    } 
    return 0; 

 

补充:软件开发 , C++ ,
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