Problem:
Score : 300 points

Problem Statement

Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?

Constraints

1 \leq a, b, c \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

a \ b \ c

Output

If \sqrt{a} + \sqrt{b} < \sqrt{c}, print Yes; otherwise, print No.

Sample Input 1

2 3 9

Sample Output 1

No

\sqrt{2} + \sqrt{3} < \sqrt{9} does not hold.

Sample Input 2

2 3 10

Sample Output 2

Yes

\sqrt{2} + \sqrt{3} < \sqrt{10} holds.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<vector>
#include<tgmath.h>
using namespace std;

long double EPS = 1E-14;

// r の(誤差付きの)符号に従って, -1, 0, 1 を返す.
int sgn(long double r){ return (r > EPS) - (r < -EPS); }
// a, b の(誤差付きの)大小比較の結果に従って, -1, 0, 1 を返す.
int sgn(long double a, long double b){ return sgn(a - b); }

int main(){
    long double a,b,c;
    cin >> a >> b >> c;
    if(sgn(sqrtl(c),sqrtl(a)+sqrtl(b)) == 1){
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }
}