HDU 2199 Can you solve this equation?(二分搜索)

题目链接

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!

 

题解：二分搜索，精度见代码。一直WA，找不到错，最后发现是少了一个感叹号！坑啊。

判断无解利用分析这个函数的导函数在[0，100]上恒大于0，所以这个函数在[0，100]上单调递增的性质来判断。

 

#include <cstdio>
#include <iostream>
#include <string>
#include <sstream>
#include <cstring>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <map>
#define ms(a) memset(a,0,sizeof(a))
#define msp memset(mp,0,sizeof(mp))
#define msv memset(vis,0,sizeof(vis))
using namespace std;
#define LOCAL
int y;
double fun(double x)
{
    return 8*pow(x,4.0)+7*pow(x,3.0)+2*pow(x,2.0)+3*x+6;
}
void solve()
{
    if(fun(0)>y||fun(100)<y)
    {
        printf("No solution!
");
        return;
    }
    double a=0,b=100,ans,m;
    while(b-a>1e-6)
    {
        m=(a+b)/2;
        ans=fun(m);
        if(ans>y)b=m-1e-7;
        else a=m+1e-7;
    }
    m=(a+b)/2.0;
    printf("%.4lf
",m);
    return;
}
int main()
{
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
#endif // LOCAL
    //Start
    int N;
    cin>>N;
    while(N--)
    {
        cin>>y;
        solve();
    }
    return 0;
}