MiYu原創(chuàng), 轉(zhuǎn)帖請注明 : 轉(zhuǎn)載自 ______________白白の屋
題目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2899
題目描述:
Strange fuction
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 95 Accepted Submission(s): 75
Problem Description
Now, here is a fuction:
F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
Can you find the minimum value when x is between 0 and 100.
Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has only one real numbers Y.(0 < Y <1e10)
Output
Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.
Sample Input
2
100
200
Sample Output
-74.4291
-178.8534
題目分析:
純數(shù)學(xué)題, 需要微積分的知識分析題目...........我承認(rèn)....我數(shù)學(xué)沒怎么學(xué)好..........
分析出來使用2分搜索就可以了..........
代碼如下:
#include <iostream>
#include <cmath>
using namespace std;
#define POW(x) ( (x) * (x) )
#define POW3(x) ( POW(x) * (x) )
MiYu原創(chuàng), 轉(zhuǎn)帖請注明 : 轉(zhuǎn)載自 ______________白白の屋
#define POW4(x) ( POW(x) * POW(x) )
double y = 0;
double cal ( double n )
{
return 8.0 * POW4(n) + 7 * POW3(n) + 2 * POW(n) + 3 * n + 6 ;
}
int main ()
{
int T;
scanf ( "%d",&T );
while ( T -- )
{
scanf ( "%lf",&y );
if ( cal(0) > y || cal(100) < y )
{
printf ( "No solution!\n" );
continue;
}
double l = 0.0, r = 100.0,res = 0.0;
while ( r - l > 1e-6 )
{
double mid = ( l + r ) / 2.0;
res = cal ( mid );
if ( res > y )
r = mid - 1e-6;
else
l = mid + 1e-6;
}
printf ( "%.4lf\n",( l + r ) / 2.0 );
}
return 0;
}