HDU 2199 Can you solve this equation（二分答案）

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 == 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

【题意】

在[0,100]内找一个方程的实数解。

【分析】

求导易得函数单增，只需二分即可。

注意精度！

【代码】

　

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
#define debug(x) cerr<<#x<<" "<<x<<'\n';
using namespace std;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const int N=1e5+5;
int n;double Y;
inline double f(double x){
	return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;
}
int main(){
	for(n=read();n--;){
		scanf("%lf",&Y);
		if(Y<f(0)||Y>f(100)){puts("No solution!");continue;}
		double l=0,r=100,mid,ans;
		while(r-l>1e-6){
			mid=(l+r)/2.0;
			if(Y-f(mid)<=1e-5){
				ans=mid;
				r=mid;
			}
			else{
				l=mid;
			}
		}
		printf("%.4lf\n",ans);
	}
	return 0;
}