ACM:HDU 2199 Can you solve this equation? 解题报告 -二分、三分

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16281    Accepted Submission(s): 7206


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
 

数学题，直接二分，代码

#include"iostream"
#include"algorithm"
#include"cstdio"
#include"cstring"
#include"cmath"
#define max(a,b) a>b?a:b
#define min(a,b) a<b?a:b
#define MX 10000 + 50
using namespace std;

double f(double x)
{
    return  8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6;
}

int main()
{
    int n;
    double m;
    while(~scanf("%d",&n))
    {
        for(int k=1; k<=n; k++)
        {
            scanf("%lf",&m);

            double i=0.0;
            if(m<f(0)||m>f(100))
            {
                printf("No solution!\n");continue;
            }
            double fr=0.0,ed=100.0;
            for(; fabs(f(fr)-f(ed))>1e-4;)
            {
                i=(fr+ed)/2;
                if(f(i)<m)fr=i;
                else ed=i;
            }
            {
                printf("%.4lf\n",i);
            }
        }
    }
    return 0;
}