【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\n",ans);
else
puts("-1");
return 0;
}