Codeforces Round #320 (Div. 2) C. A Problem about Polyline

C. A Problem about Polyline

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – ....

We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.

Input

Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).

Output

Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output  - 1 as the answer.

Sample test(s)

input

3 1

output

1.000000000000

input

1 3

output

-1

input

4 1

output

1.250000000000

题目大意：

有一根折线经过 (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – ..

输入（a,b）这根折线经过（a,b）输出最大的k的值

解法：二分求解

#include "iostream"#include "string.h"#include "string"#include "queue"#include "cstdio"using namespace std; double a,b;int main(int argc, char* argv[]){    cin>>a>>b;    int left=1,right=1000000001;    if(a<b)        puts("-1");    else    {        double ans=0;        double tempy=(a+b)/2;        while(left<=right)        {            int mid=(left+right)/2;            if(tempy/mid>=b)                left=mid+1;            else                right=mid-1;        }        printf("%.11lf\n",(double)tempy/right);    }    return 0;}