HDU--杭电--Can you solve this equation?--二分

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6176    Accepted Submission(s): 2889

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!

 

 
题意就是看能不能找到一个 Y = 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6
有就输出，没有就输出那句英文
思路：题目说了从0到100中找，所以直接进行二分就是了

#include <iostream>#include <stdio.h>using namespace std;double f(double x){    return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;}int main (void){    int n;    double y,first,last,d;    cin>>n;    while(cin>>y)    {        if(y<f(0)||y>f(100))//不再范围内的直接特解掉        {            puts("No solution!");            continue;        }        first=0;last=100;        while(last-first>1e-8)//二分查找        {            d=(last+first)/2;            if(f(d)<y)            {                first=d+1e-8;//1e-8就是-1/100000000，这是对付精度的            }            else            {                last=d-1e-8;            }        }printf("%.4lf\n",d);    }    return 0;}