HDU

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20612    Accepted Submission(s): 9045

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

2100-4

 

Sample Output

1.6152No solution!

 

很简单的题死活WA了好几次……首先是输入的x可能为浮点型，其次是函数是单调递增且有取值范围，超过最值则输出没有结果，最后是计算精度，虽然是保留4位但是还是要计算到1e-8以内……

#include <iostream>#include<stdio.h>#include<math.h>using namespace std;int main(){    int T;    double x,y,z,l,h;    cin>>T;    while(T--){        cin>>y;        if(y<6||y>807020306){            cout<<"No solution!"<<endl;            continue;        }        x=50.0;        h=100.0;        l=0.0;        while(fabs(h-l)>1e-8){            z=8.0*x*x*x*x + 7.0*x*x*x + 2.0*x*x + 3.0*x + 6.0;            if(z>y){                h=x;            }else{                l=x;            }            x=(h+l)/2;        }        //printf("%.5lf\n",x);        printf("%.4lf\n",x);    }    return 0;}