Can you solve this equation?

                 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                      Total Submission(s): 25477    Accepted Submission(s): 10963

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!
 

http://acm.hdu.edu.cn/showproblem.php?pid=2199

題意：

給你一個Y，讓你求解8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y，且x的範圍為[0,100]。如果有解，則輸出x的值，沒有解，則輸出No solution!

思路：

二分的一個簡單題，判斷左邊是否大於右邊，如果大於，則r = mid，如果小於，則l = mid，然後遞迴呼叫。

程式碼：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
double f(double mid)
{
	return ((8.0 * mid * mid * mid * mid) + ((7.0 * mid * mid * mid)) + ((2.0 * mid * mid)) + 3.0 * mid +  6);
}
double con(double l, double r, double r1) {
	double mid = (l+r) / 2.0;
	
	if(r-l>0.000000001) {
		if(f(mid) == r1)
			return mid;	
		if(r1 > f(mid)) {
			return con(mid, r, r1);
		}
		if(f(mid) > r1)
			return con(l, mid, r1);
		
	}
	
	return mid;
	 
}
int main()
{
	int t;
	cin >> t;
	while(t--) {
		long long y;
		scanf("%lld", &y);
		if(y < 6 || y > f(100)) {
			cout << "No solution!" << endl;
		}
		else {
			double l = 0, r = 100; 
			double r1 = (double) y;
			
			printf("%0.4llf\n",con(l,r,r1));
		}	
	} 
	
	return 0;
 }