hud 2199 Can you solve this equation?(二分搜索）

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!
        
        
        
        

 

题目大意：根据题目所给定的公式，和后面给的一个Y值，在0~100的区间上求x的值。

解题思路：将公式求导可以证明曲线在0~100这个区间上是单调递增的，所以可以用二分搜索法去做，因为做这道题目的时候实在训练赛的时候，而且以前没有写过二分的题目，所以没想太多，直接写了一个递归的二分搜索，其实只要用一个whlie循环就可以搞定。

#include <stdio.h>
#include <math.h>

int t, y;

long long count(double x){
    double sum = 8 * pow(x, 4) + 7 * pow(x, 3) + 2 * pow(x, 2) + 3 * x + 6 - y;
   return floor(sum * 10000);
} 

void dfs(double a, double b){
    long long  x = count((a + b) / 2.0);
    
    if (x == 0)
	printf("%.4lf\n", (a + b) / 2.0);
    else if (x > 0)
	dfs(a, (a + b) / 2.0);
    else
	dfs((a + b) / 2.0, b);
}

int main(){
    scanf("%d", &t);
    while (t--){
	scanf("%d", &y);
	long long a = count(0);
	long long b = count(100);

	if (a == 0)
	    printf("0.0000\n");
	else if (b == 0)
	    printf("100.0000\n");
	else if (a > 0 && b < 0)
	    dfs(100, 0);
	else if (a < 0 && b > 0)
	    dfs(0, 100);
	else
	    printf("No solution!\n");
    }
    return 0;
}