HDU 3910 (13.10.31)

Description

Maybe you know “San Guo Sha”, but I guess you didn’t hear the game: “Liang Guo Sha”!

Let me introduce this game to you. Unlike “San Guo Sha” with its complicated rules, “Liang Guo Sha” is a simple game, it consists only four cards, two cards named “Sha”, and the other named “Shan”.

Alice and Bob are good friends, and they’re playing “Liang Guo Sha” now.
Everyone has two cards: a “Sha” and a “Shan”. Each round, everyone choose a card of his/her own, and show it together(Just show the selected card, do not need to put it away). If both of them choose “Sha”,
then Alice gets A points,
and Bob loses A points; if both of them choose “Shan”,
then Alice gets B points,
and Bob loses B points; otherwise,
Bob gets C points,
and Alice loses C points.

Both Alice and Bob wants to get points as many as possible, they thought a optimal strategy: Calculating a percentage of choosing card “Sha” in order to ensure that even the opponent uses the optimal strategy, he/she can still get a highest point exceptation.

  

Here is the question, if both Alice and Bob use the optimal strategy to make their points higher, what is the expectation point which Alice can get in a round?

 

Input

Several test case, process to EOF.

  Each test case has only a line, consists three positive integers: A, B, C respectively.

  1 <= A, B, C <= 100000

 

Output

Each test case just need to output one line, the expectation point that Alice can get. Round to 6 decimal points.

 

Sample Input

2 10 4
3 3 3

 

Sample Output

0.200000
0.000000

Hint

In test case 1, both Alice and Bob calculated the best percentage of choosing “Sha”, and the their percentage are the same: 70%.
If Bob do not choose the best percentage, his strategy might be targetd.

For example, if Bob choose 100%, then Alice can change her percentage to 100%, Bob might lose many points. Bob is clever, so he won’t do that.

参考题解：http://blog.csdn.net/ivan_zjj/article/details/7927970

AC代码

#include<stdio.h>

int main() {
    double a, b, c;
    while(scanf("%lf %lf %lf", &a, &b, &c) == 3) {
        double ans = (a*b - c*c) / (a + b + 2*c);
        printf("%.6lf\n", ans);
    }
    return 0;
}