| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 12160 | Accepted: 3160 |
Description
When a thin rod of length L is heated n degrees, it expands to a new length
Your task is to compute the distance by which the center of the rod is displaced.
Input
Output
Sample Input
1000 100 0.0001 15000 10 0.00006 10 0 0.001 -1 -1 -1
Sample Output
61.329 225.020 0.000, where C is the coefficient of heat expansion.
When a thin rod is mounted on two solid walls and then heated, it expands and takes the shape of a circular segment, the original rod being the chord of the segment.
Your task is to compute the distance by which the center of the rod is displaced.
Input
Output
Sample Input
1000 100 0.0001 15000 10 0.00006 10 0 0.001 -1 -1 -1
Sample Output
61.329 225.020 0.000
题目大意:有已知长度的一根木棒,随着温度变化有伸缩变化,根据公式L'=(1+n*C)*L得到新的木棍长度,在两面墙之间会弯曲,求木棍前后状态的中点之间的距离。
解题思路:联系相关几何知识得到公式:L'=2*r*arcsin(L/(2*r),r=(4*h*h+L*L)/(8*h)。采用二分枚举h的值得到r的值,然后公式1作为判断条件,直接求出h的值。
解题过程:wa点:注意浮点型的二分结束循环的条件是(high-low>minx),这题minx精度尽量大一些才能过。
代码如下:
#include <cstdio>
#include <cmath>
const double minx=1e-8;
int main()
{
double l,n,c,L,high,low,mid,r;
while(scanf("%lf%lf%lf",&l,&n,&c)!=EOF&&(l>=0||n>=0||c>=0))
{
L=(1+n*c)*l;
high=l/2;
low=0.0;
while(high-low>minx)
{
mid=(high+low)/2;
r=(4*mid*mid+l*l)/(8*mid);
if(2*r*asin(l/(2*r))<L)
low=mid;
else
high=mid;
}
printf("%.3f\n",mid);
}
return 0;
}
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