杭电 HDU ACM 2199 Can you solve this equation?
发布时间：2020-12-13 20:09:11 所属栏目：PHP教程 来源：网络整理
导读：Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 11180Accepted Submission(s): 5151 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): 11180 Accepted Submission(s): 5151 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
Sample Output
Author
Redow
搜索题，嗯， 这次知道了有个搜索名次，2分搜索。这样查找答案确切感觉特别妙。low 和high 无穷夹逼答案。所以由于最后结果只需满足1定的精确度。这样可以找到适合的条件跳出while。
唉，如果不知道这样的话，浮点数暴力简直无语了都。
#include<iostream>
#include<cmath>
#include<stdio.h>
using namespace std;
double A(double x)
{
return (8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*pow(x,1)+6);
}
int main()
{
double y;
int T;
cin>>T;
while(T--)
{
cin>>y;
if(A(0)>y||A(100)<y)
{
cout<<"No solution!"<<endl;
continue;
}
else{
double high,low,mid;
high=100;
low=0;
while(low+1e⑻<high)
{
mid=(high+low)/2;
if(A(mid)>y)
high=mid;
else
low=mid;
}
printf("%.4lf
",low);
}
}
return 0;
}
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