HDU 2199 Can you solve this equation? 二分

主要是练习下二分思想+浮点数精度控制

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7029    Accepted Submission(s): 3266

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!
   
   
   
   

 

这个方程的单调性是很好证的，不断求导很容易证得，在单调内求解用二分，主要是尝试了下各种精度控制。

#include <cstdio>
#include <cmath>
#include <cstdlib>

const double eps = 1e-7;

double cal(double x){
	return 8*pow(x,4.0)+7*pow(x,3.0)+2*pow(x,2.0)+3*x+6.0;
}

int main(){
	int t;
	double low,high,mid,y,res;

	scanf("%d",&t);
	while(t--){
		scanf("%lf",&y);
		if(y<cal(0.0) || y>cal(100.0)){
			printf("No solution!\n");
			continue;
		}
		low = 0.0;
		high = 100.0;
		while(high-low>eps){
			mid = (low+high)/2;
			res = cal(mid);
			if(res<y){
				low = mid + 1e-7;
			}else{
				high = mid - 1e-7;
			}
		}
		printf("%0.4lf\n",mid);
	}
	return 0;
}