HDU 2199 Can you solve this equation?（二分法求近似解）

来源：互联网发布：守望先锋手机数据查询编辑：程序博客网时间：2024/06/04 18:58

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 16588 Accepted Submission(s): 7366

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

2100-4

Sample Output

1.6152No solution!

Author

Redow

思路：简单的二分求解方程，但是有助于理解二分的含义。

代码：

#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<algorithm>#define MYDD 11030using namespace std;typedef double DO;//定义 DO 数据类型 DO fun(DO x) {return 8.0*x*x*x*x+7*x*x*x+2*x*x+3*x+6;}//8* X^4+ 7* X^3+ 2* X^2+ 3 * X +6= Yint main() {int t;scanf("%d",&t);while(t--) {DO y;scanf("%lf",&y);if(y<fun(0)||fun(100)<y) {//函数的最大最小值 puts("No solution!");continue;}int v=64;DO turn=0.0,right=100.0,middle;while(v--) {middle=(turn+right)/2.0;if(fun(middle)>y) {right=middle;} else {turn=middle;}}printf("%.4lf\n",middle);}return 0;}

0 0