Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5999    Accepted Submission(s): 2828 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,…

HDU 2199 Can you solve this equation?     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. InputThe first line of the input contains an integer T(1<=T<=100) which mea…

http://acm.hdu.edu.cn/howproblem.php?pid=2199 Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13468    Accepted Submission(s): 6006 Problem Description Now,given the…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=2199 Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 25039    Accepted Submission(s): 10776 Problem Description Now,gi…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7493    Accepted Submission(s): 3484 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,…

Can you solve this equation? Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Now,given the equation *x^ + *x^ + *x^ + *x + == Y,can you find its solution between and ; No…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7156    Accepted Submission(s): 3318 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y…

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 solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 27728    Accepted Submission(s): 11717 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 ==…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 22742    Accepted Submission(s): 9865 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 ==…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6763    Accepted Submission(s): 3154 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,…

解题思路:给出一个方程 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,求方程的解. 首先判断方程是否有解,因为该函数在实数范围内是连续的,所以只需使y的值满足f(0)<=y<=f(100),就一定能找到该方程的解,否则就无解. 然后是求解过程, 假设一个区间[a,b],mid=(a+b)/2,如果f(a)*f(b)<0,那么函数f(x)在区间[a,b]至少存在一个零点,如果f(a)<0,说明0点在其右侧,那么将a的值更新为当前mid的值,如果f(a)&g…

题意 给Y值,找到多项式 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y 在0到100之间的解. 思路 从0到100,多项式是单调的,故用二分法求解. 代码 double calc(double x){ return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6; } int main(){ int T; cin>>T; while(T--){ double Y; cin>>Y; double L,R; L = 0.0, R= 100.0;…

Now,given the equation 8x^4 + 7x^3 + 2x^2 + 3x + 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 fol…

#include<stdio.h> #include<math.h> double f(double x) { return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6; } int main(void) { int t; double y,x1,x2,x3,y1,y2,y3; scanf("%d",&t); while(t--) { scanf("%lf",&y); x1=0; x2=100; y1=f(x1…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12766    Accepted Submission(s): 5696 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y…

Problem Description Now,given the equation 8x^4 + 7x^3 + 2x^2 + 3x + 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 cas…

题意:给出一个数让你求出等于这个数的x 策略:如题. 由于整个式子是单调递增的.所以能够用二分. 要注意到精度. 代码: #include <stdio.h> #include <string.h> #include <math.h> #define eps 1e-10 #define f(x) 8*pow(x, 4) + 7*pow(x, 3) + 2*pow(x, 2) + 3*x int main() { int t; double n; scanf("…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=2199 Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8595    Accepted Submission(s): 3957 Problem Description Now,given…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5667    Accepted Submission(s): 2681http://acm.hdu.edu.cn/showproblem.php?pid=2199 Problem Description Now,given the…

Can you solve this equation? Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 61   Accepted Submission(s) : 37 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,can y…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1182 Accepted Submission(s): 558   Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,can…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 25633    Accepted Submission(s): 11018 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 ==…

Solve a given equation and return the value of x in the form of string "x=#value". The equation contains only '+', '-' operation, the variable x and its coefficient. If there is no solution for the equation, return "No solution". If th…

D - Can you solve this equation? 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. InputThe first line of the input contains an integer T(1<=T<=100) which means the num…

[LeetCode]640. Solve the Equation 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.me/ 题目地址:https://leetcode.com/problems/solve-the-equation/description/ 题目描述: Solve a given equation and return the value of x in the…

Birthday Toy Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 644    Accepted Submission(s): 326 Problem Description AekdyCoin loves toys. It is AekdyCoin’s Birthday today and he gets a special “…

+二分法求平方根 x = float(raw_input('Enter the number')) low = 0 high = x guess = (low + high ) / 2 if x < 0: print 'Number Error' while abs(guess**2 - x) > 1e-5: if guess**2 < x: low = guess else: high = guess guess = (low + high) / 2 print 'The root o…

一:用迭代法求 x=√a.求平方根的迭代公式为:X(n+1)=(Xn+a/Xn) /2. #define _CRT_SECURE_NO_WARNINGS #include <stdio.h> #include <stdlib.h> #include <math.h> int main() { double x1, x2; float a; scanf("%f", &a); x2 = 1.0; do { x1 = x2; x2 = (x1 +…

03-树1. 二分法求多项式单根(20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 杨起帆(浙江大学城市学院) 二分法求函数根的原理为:如果连续函数f(x)在区间[a, b]的两个端点取值异号,即f(a)f(b)<0,则它在这个区间内至少存在1个根r,即f(r)=0. 二分法的步骤为: 检查区间长度,如果小于给定阈值,则停止,输出区间中点(a+b)/2:否则 如果f(a)f(b)<0,则计算中点的值f((a+b)/2):…