Can you solve this equation?（二分查找的简单应用）

Can you solve this equation?

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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!

思路：
记f(x)为等式左边的值，如果f(100.0)小于Y或f(0.0)大于Y，无解，否则必有解，且解的范围为[0.0, 100.0],不断二分区间，直至找到f(x) = Y

 1 #include <iostream>
 2 #include <string>
 3 #include <cstdio>
 4 #include <cmath>
 5 #include <cstring>
 6 #include <algorithm>
 7 #include <map>
 8 #include <vector>
 9 #include <queue>
10 #include <stack>
11 #define LL long long
12 #define MAXI 2147483647
13 #define MAXL 9223372036854775807
14 #define eps (1e-8)
15 #define dg(i) cout << "*" << i << endl;
16 
17 using namespace std;
18 
19 double Solve(double x)
20 {
21     return (8*x*x*x*x + 7*x*x*x + 2*x*x + 3*x + 6);
22 }
23 
24 int main()
25 {
26     int t;
27     double y, low, high, mid;
28     scanf("%d", &t);
29     while(t--)
30     {
31         scanf("%lf",&y);
32         low = 0.0;
33         high = 100.0;
34         if(y > Solve(100.0) || y < Solve(0.0)) puts("No solution!");
35         else
36         {
37             while(low + eps < high)
38             {
39                 mid = (low + high) * 0.5;
40                 if(y > Solve(mid) + eps) low = mid;
41                 else if(y < Solve(mid) - eps) high = mid;
42                 else break;
43             }
44             printf("%.4lf\n", mid);
45         }
46     }
47     return 0;
48 }