(hdu step 4.1.1)Can you solve this equation?(二分法を用いて方程式群の解を解く)

タイトル:
Can you solve this equation?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 915 Accepted Submission(s): 436
 
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!

 
Author
Redow
 
 
Recommend
lcy
テーマ分析:
               簡単な問題.二分法を用いて方程式群の解を求める.二分法を用いて方程式群の解を求める前提は,この方程式が単調性を有することである.
コードは次のとおりです.

/*
 * a.cpp
 *
 *  Created on: 2015 2 16 
 *      Author: Administrator
 */

#include <iostream>
#include <cstdio>
#include <cmath>



using namespace std;

const double delta = 1e-6;//1e10  10^10

double function(double x){
	return 8*pow(x,4.0) + 7*pow(x,3) + 2*pow(x,2) + 3*x + 6;
}



/**
 *             
 * y:       
 */
double work(double y){
	double mid;
	double left = 0;
	double right = 100;

	while(right - left > delta){//               
		mid = (left+right)/2;//     

		if(function(mid) < y){//  y     ...      
			left = mid+(1e-7);//       。                     
		}else{//  
			right = mid-(1e-7);//       
		}
	}

	return (left+right)/2;//      
}




int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		double y;
//		scanf("%d",&y);

		cin >> y;

		if(function(0) <= y && y <= function(100)){
			printf("%.4lf
",work(y));
//			printf("%.4lf
",find(y));
		}else{
			printf("No solution!
");
		}
	}

	return 0;
}