杭电 2899 数学 牛顿迭代法

   第一次写这样的题目，写出来很有成就感啊。用二分法和牛顿迭代法都可以解决。题目：

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 739    Accepted Submission(s): 566

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
   
   
   
   

 

牛顿迭代法ac代码：

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
double ee=1e-10;
double y;
double mi(double s,int x){
  double ss=1.00000000;
  for(int i=1;i<=x;++i)
	  ss*=s;
  return ss;
}
double f(double x){
  return 42*mi(x,6)+48*mi(x,5)+21*mi(x,2)+10*x-y;  
}
double ff(double x){
  return 252*mi(x,5)+240*mi(x,4)+42*x+10;
}
double solve(double x){
  return 6*mi(x,7)+8*mi(x,6)+7*mi(x,3)+5*mi(x,2)-y*x;
}
int main(){
  int numcase;
  scanf("%d",&numcase);
  while(numcase--){
    scanf("%lf",&y);
	double fx=f(0.00000000);
	double fy=ff(0.00000000);
	double fz=0.00000000-fx/fy;
	double ans=0.00000000;
	while(abs(fz-ans)>ee){
	  ans=fz;
	  fx=f(fz);
	  fy=ff(fz);
	  fz=fz-fx/fy;
	}
	double newans=solve(ans);
	printf("%.4lf\n",newans);
  }
  return 0;
}

二分法ac代码：

#include <iostream>
#include <cstdio>
using namespace std;
double ee=1e-10;
double mi(double s,int x){
  double ss=1.00000000;
  for(int i=1;i<=x;++i)
	  ss*=s;
  return ss;
}
double binary_search(double x){
  double ll=0.00000000,rr=100.00000000;
  double mid;
  double mm,nn;
  while(rr-ll>ee){
   mid=(ll+rr)/2.0;
   mm=42*mi(mid,6)+48*mi(mid,5)+21*mi(mid,2)+10*mid;
   nn=mm-x;
   if(nn>ee)
	   rr=mid;
   else if(nn<ee)
	   ll=mid;
   else if(nn==ee)
	   break;
  }
  return mid;
}
double solve(double x,double ss){
  return 6*mi(x,7)+8*mi(x,6)+7*mi(x,3)+5*mi(x,2)-ss*x;
}
int main(){
  int numcase;
  double y;
  scanf("%d",&numcase);
  while(numcase--){
    scanf("%lf",&y);
	double ans=binary_search(y);
	double mmax=0.0;
	if(mmax>solve(100.000,y))
		mmax=solve(100.000,y);
	if(mmax>solve(ans,y))
		mmax=solve(ans,y);
	printf("%.4lf\n",mmax);
  }
  return 0;
}