zoukankan      html  css  js  c++  java
  • HDU 2199 Can you solve this equation? (二分 水题)

    Can you solve this equation?


    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 6023    Accepted Submission(s): 2846

    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
     


    一道初学者的二分题,讲课用

    #include <iostream>
    #include <cstdio>
    using namespace std;
    
    double y;
    
    double getfx(double x){
    	return 8*x*x*x*x + 7*x*x*x + 2*x*x + 3*x + 6;
    }
    
    void computing(){
    	if(getfx(100)<y-1e-3 || getfx(0)>y+1e-3){
    		printf("No solution!
    ");
    		return;
    	}
    	double l=0,r=100;
    	while(r-l>1e-6){
    		double mid=(l+r)/2;
    		if(getfx(mid)>y) r=mid;
    		else l=mid;
    	}
    	printf("%.4lf
    ",r);
    }
    
    int main(){
    	int t;
    	scanf("%d",&t);
    	while(t-- >0){
    		scanf("%lf",&y);
    		computing();
    	}
    	return 0;
    }





  • 相关阅读:
    实现Callable接口实现多线程
    匿名内部类方式实现
    实现Runnable接口方式
    后台线程
    继承Thread类
    线程中断详解
    第六章、Linux 的档案权限不目彔配置
    安装virtual box
    CISCO实验记录六:EIGRP路由协议
    zabbix监控项整理Items-key
  • 原文地址:https://www.cnblogs.com/fuhaots2009/p/3433032.html
Copyright © 2011-2022 走看看