hdu 2199

Problem Description
Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,can you find its solution between 0 and 100;
Now please try your lucky.


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 a real number Y (fabs(Y) <= 1e10);


Output
For each test case, you should just output one real number(accurate up to 4 decimal places),which is the solution of the equation,or “No solution!”,if there is no solution for the equation between 0 and 100.


Sample Input
2
100
-4


Sample Output
1.6152

No solution!

迭代法代码：

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <iomanip>
#include <cmath>
#include <algorithm>
using namespace std;
double cmp(double low,double high,double y);
double mcp(double x);
int main()
{
	int T;
	double n,m;
	while(cin>>T){
		while(T--){
			cin>>n;
			if(n<mcp(0)||n>mcp(100))
			  cout<<"No solution!"<<endl;
		else {
		      m=cmp(0,100,n);
		      cout<<setiosflags(ios::fixed)<<setprecision(4)<<m<<endl;}  
        }
	}
	return 0;
}
double cmp(double low,double high,double y)
{
	double mid;
	while(high-low>1e-10){
		mid=(low+high)/2.0;
		if(mcp(mid)>y) high=mid;
		else low=mid;
	}
	return mid;
}
double mcp(double x){
	return 8.0*x*x*x*x+7.0*x*x*x+2.0*x*x+3.0*x+6.0;
}

递归法代码：

如果递归的判断终止条件为low<high，时间超限；因为这个类型是double型，有些数会递归很多次以至n次才会的到结果，以至超时；由于提上要求是4位小数，所以只需要让low<high-0.0000001即可；

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <iomanip>
#include <cmath>
#include <algorithm>
using namespace std;
double cmp(double low,double high,double y);
double mcp(double x);
int main()
{
	int T;
	double n,m;
	while(cin>>T){
		while(T--){
			cin>>n;
			if(n<mcp(0)||n>mcp(100))
			  cout<<"No solution!"<<endl;
		else {
		      m=cmp(0,100,n);
		      cout<<setiosflags(ios::fixed)<<setprecision(4)<<m<<endl;}  
        }
	}
	return 0;
}
double cmp(double low,double high,double y)
{
	double mid=(low+high)/2;
	if(high-low>1e-10){
	    if(mcp(mid)>y)
	        return cmp(low,mid,y);
        else if(mcp(mid)<y)
            return cmp(mid,high,y);
        else return mid;
	}
	return mid;
}
double mcp(double x){
	return 8.0*x*x*x*x+7.0*x*x*x+2.0*x*x+3.0*x+6.0;
}

思路：先要判断这个函数的单调性，之后在就容易多了，给你一个数y；求解出对应的x，如果x是在0-100之间输出x浮点数为4；否则输出“No solution!”；

说白了就是把y移到等式左边是f(x)= 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 -Y；求是零点的值；如果x是在0-100之间输出x，浮点数为4；否则输出“No solution!”；