【26.09%】【codeforces 579C】A Problem about Polyline

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

Examples
input
3 1
output
1.000000000000
input
1 3
output
-1
input
4 1
output
1.250000000000
Note
You can see following graphs for sample 1 and sample 3.

【题目链接】:http://codeforces.com/contest/579/problem/C

【题解】

先把两种类型的线段方程搞出来:
/型的为y=x-2k*x0
型的为y=-x+2k*x0
先考虑/型
移项一下
x0 = (x-y)/(2*k);
然后把(a,b)代入
x0 = (a-b)/(2*k); ·····①
然后再对k考虑
k = (a-b)/(2*x0)
因为x0为整个图像的最高的的y坐标;
所以x0 >= b;
则x0有最小值
则k<=(a-b)/(2*b);
再带回①式
因为k有最大值;所以
x0 >=(a-b)/(2*[(a-b)/(2*b)]);
[x]表示不超过x的最大整数;
对型的直线同理可以得到
x0 >=(a+b)/(2*[(a+b)/(2*b)]);
要注意(a+b)/(2*b)和(a-b)/(2*b)都要大于等于1才行;
因为k就是原题中的n;而n显然应该要大于0;

【完整代码】

#include <bits/stdc++.h>
#define LL long long

using namespace std;

LL a,b;

int main()
{
    cin >> a >> b;
    double ans;
    bool flag1 = true,flag2 = true;
    if (a>=b)
    {
        if (a==b)
            ans = b;
        else
        {
            int temp = (a-b)/(2.0*b);
            if (temp <1)
                flag1 = false;
            ans =(a-b)/(2*temp*1.0);
        }
    }
    else
        flag1 = false;
    int temp = (a+b)/(2.0*b);
    if (temp < 1)
        flag2 = false;
    {
        double temp1 = (a+b)/(2*temp*1.0);
        if (!flag1)
            ans = temp1;
        else
            ans = min(ans,temp1);
    }
    if (flag1 || flag2)
        printf("%.12lf
",ans);
    else
        puts("-1");
    return 0;
}