反思

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

int main(){
	int a,b,c1,c2;
	double ans,x,y;
	scanf("%d %d", &a, &b);
	if(a<b){
		puts("-1");
	}
	else if(a==b){
		printf("%d\n", a);
	}else{
		double x = a - b, y = a + b;
        int c1 = x/b/2, c2 = y/b/2;
        ans = min(x / c1 / 2, y / c2 / 2);
        printf("%.10lf\n", ans);

	}
	
	return 0;
}

 

 

没读懂题意 而且。。。 写的错误不堪。。

注意反思

 A Problem about Polyline

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

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 Input

Input

3 1

Output

1.000000000000

Input

1 3

Output

-1

Input

4 1

Output

1.250000000000

Hint

You can see following graphs for sample 1 and sample 3.

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
    double x;
    double a,b;
    while(cin>>a>>b)
    {
        if(b>a)
        {
            x=-1;
        }
        else if(b==a)
        {
            x=a;
        }
        else
        {
            double r=(a+b)/2;
            double d=(a-b)/2;
            //cout<<r<<" "<<d<<endl;
            int rr=r/b,dd=d/b;
            while(r/rr<b)
            {
                rr--;
            }
            while(d/dd<b)
            {
                dd--;
            }
            //cout<<rr<<" "<<dd<<endl;
            if((double)r/rr>=(double)d/dd)
            {
                x=(double)d/dd;
            }

            else
            {
                x=(double)r/rr;
            }
        }
        printf("%.12f\n",x);
    }

    return 0;
}