【UVa】10566 - Crossed Ladders(二分 & 数学)
2016-07-26 12:19
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Crossed Ladders
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit Status
Description
A narrow street is lined with tall buildings. An x foot long ladder is rested at the base of the building on the right side of the street and leans on the building on the left side. A y foot long ladder is rested at the
base of the building on the left side of the street and leans on the building on the right side. The point where the two ladders cross is exactly c feet from the ground. How wide is the street?
Input
Input starts with an integer T (≤ 10), denoting the number of test cases.
Each test case contains three positive floating point numbers giving the values of x, y, and c.
Output
For each case, output the case number and the width of the street in feet. Errors less than 10-6 will be ignored.
Sample Input
4
30 40 10
12.619429 8.163332 3
10 10 3
10 10 1
Sample Output
Case 1: 26.0328775442
Case 2: 6.99999923
Case 3: 8
Case 4: 9.797958971
纯纯的数学题,设底边为w,左小边为a,列出方程解出来,然后用二分求解。
代码如下:
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
double x,y,c;
double check(double w)
{
return 1 - c / sqrt(x * x - w * w) - c / sqrt(y * y - w * w);
}
int main()
{
double l,r,mid;
int u;
int ant = 1;
scanf ("%d",&u);
while (u--)
{
scanf ("%lf %lf %lf",&x,&y,&c);
printf ("Case %d: ",ant++);
l = 0;
r = min(x,y);
while (r - l > 1e-9)
{
mid = (l + r) / 2;
if (check(mid) > 0)
l = mid;
else
r = mid;
}
printf ("%.6lf\n",l);
}
return 0;
}
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit Status
Description
A narrow street is lined with tall buildings. An x foot long ladder is rested at the base of the building on the right side of the street and leans on the building on the left side. A y foot long ladder is rested at the
base of the building on the left side of the street and leans on the building on the right side. The point where the two ladders cross is exactly c feet from the ground. How wide is the street?
Input
Input starts with an integer T (≤ 10), denoting the number of test cases.
Each test case contains three positive floating point numbers giving the values of x, y, and c.
Output
For each case, output the case number and the width of the street in feet. Errors less than 10-6 will be ignored.
Sample Input
4
30 40 10
12.619429 8.163332 3
10 10 3
10 10 1
Sample Output
Case 1: 26.0328775442
Case 2: 6.99999923
Case 3: 8
Case 4: 9.797958971
纯纯的数学题,设底边为w,左小边为a,列出方程解出来,然后用二分求解。
代码如下:
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
double x,y,c;
double check(double w)
{
return 1 - c / sqrt(x * x - w * w) - c / sqrt(y * y - w * w);
}
int main()
{
double l,r,mid;
int u;
int ant = 1;
scanf ("%d",&u);
while (u--)
{
scanf ("%lf %lf %lf",&x,&y,&c);
printf ("Case %d: ",ant++);
l = 0;
r = min(x,y);
while (r - l > 1e-9)
{
mid = (l + r) / 2;
if (check(mid) > 0)
l = mid;
else
r = mid;
}
printf ("%.6lf\n",l);
}
return 0;
}
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