Panasonic Programming Contest 2020 C (Sqrt Inequality) 题解

General idea

Enter three integers $a$，$b$，$c$ if$\sqrt{a}+\sqrt{b}<\sqrt{c}$If established, output Yes, otherwise output No.

Sample

Enter #1

2 3 9

Output #1

No

$\sqrt{2}+\sqrt{3}<\sqrt{9}$ invalid.

Input #2

2 3 10

Output #2

Yes

$\sqrt{2}+\sqrt{3}<\sqrt{1}0\; is$ established.

analysis

Wrong idea

First of all, due to sqrtthe floating-point precision error of the system function, the following code will obviously WA:

#include <cstdio>
#include <cmath>
using namespace std;

int main(int argc, char** argv)
{
    
    
	int a, b, c;
	scanf("%d%d%d", &a, &b, &c);
	double d = sqrt(double(a)) + sqrt(double(b));
	puts(d * d < c? "Yes": "No");
	return 0;
}

Therefore, this question must require special thinking ! ! !

Correct thinking

The following is the derivation process of the correct method:
$\sqrt{a}+\sqrt{b}<\sqrt{c}$
$(\sqrt{a}+\sqrt{b}{)}^2<(\sqrt{c}{)}^2$
$a+b+2\sqrt{a}b<c$
$2\sqrt{a}b<c-a-b$
$(2\sqrt{a}b{)}^2<(c-a-b{)}^2$
$4ab<(c-a-b{)}^2$
Note: There is another case, that is,$c-a-b<0$或$c<a+b$ , the answer should be yesNo. In this caseWA, themeeting is not considered, because$(c-a-b{)}^2$ will "ignore negative numbers directly"!

Code

#include <cstdio>
using namespace std;

int main(int argc, char** argv)
{
    
    
	long long a, b, c;
	scanf("%lld%lld%lld", &a, &b, &c);
	long long d = c - a - b;
	if(d < 0) puts("No"); // 特殊情况c - a - b < 0直接输出No
	else puts((d * d > 4LL * a * b)? "Yes": "No");
	return 0;
}