hdu Strange fuction

简介:

Strange fuction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 108 Accepted Submission(s): 96
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
分析:
求函数的最小值,首先求导的导函数为:G(x) = 42 * x^6+48*x^5+21*x^2+10*x-y (0 <= x <=100)
分析导函数的,导函数为一个单调递增的函数。如果导函数的最大值小于0,那么原函数在区间内单调递减。
即F(100)最小;如果但函数的最小值大于0,那么原函数在区间内单调递增,即F(0)最小。如果导函数既有正又有负
又由于导函数是单增函数,所以必有先负后正,即原函数必有先减后增的性质。求出导函数的零点就是原函数的最小值点。
求导函数最小值方法是2分法.
复制代码
#include <iostream>
#include
<stdio.h>
#include
<math.h>
using namespace std;
double f(double x,double y)
{
return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;
}
double func(double x, double y)
{
return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y;
}
int main()
{
int t;
double y,left,right,mid,ans1,ans2,ans3;
scanf(
"%d",&t);
while(t--)
{
scanf(
"%lf",&y);
if(func(0.0,y)>=0)
printf(
"%.4f\n",f(0.0,y));
else if(func(100.0,y)<=0)
printf(
"%.4f\n",f(100.0,y));
else
{
left
= 0.0;
right
= 100.0;
mid
= 50.0;
ans1
= func(left,y);
ans2
= func(mid,y);
ans3
= func(right,y);
while(fabs(ans1-ans2)>0.0001)
{
if(ans2>0)
{
ans3
= ans2;
right
= mid;
mid
= (left+right)/2;
}
else
{
ans1
= ans2;
left
= mid;
mid
= (left+right)/2;
}
ans2
= func(mid,y);
}
printf(
"%.4f\n",f(mid,y));
}
}
return 0;
}
复制代码
本文转自NewPanderKing51CTO博客,原文链接: http://www.cnblogs.com/newpanderking/archive/2011/08/24/2152492.html  ,如需转载请自行联系原作者
相关文章
|
6月前
Strange fuction(HDU--2899)
Strange fuction(HDU--2899)
|
6月前
Knight Moves(POJ2243)
Knight Moves(POJ2243)
UVa389 - Basically Speaking
UVa389 - Basically Speaking
36 0
hdu-1098 Ignatius's puzzle(费马小定理)
hdu-1098 Ignatius's puzzle(费马小定理)
154 0
hdu-1098 Ignatius's puzzle(费马小定理)
|
测试技术
HDU-1026,Ignatius and the Princess I(BFS+打印路径)
HDU-1026,Ignatius and the Princess I(BFS+打印路径)
HDU-1027,Ignatius and the Princess II
HDU-1027,Ignatius and the Princess II
HDU-1017,A Mathematical Curiosity
HDU-1017,A Mathematical Curiosity
ZOJ - Summer 2018 - Contest 1 by SBconscious - Problems - 1001: Saber
ZOJ - Summer 2018 - Contest 1 by SBconscious - Problems - 1001: Saber
94 0
|
Java
HDOJ(HDU) 2164 Rock, Paper, or Scissors?
HDOJ(HDU) 2164 Rock, Paper, or Scissors?
114 0