HDU 2199 Can you solve this equation?（二分）

原题链接

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!

题目大意

给定Y,解出X。

解题思路

由于函数是单调的，所以只需二分找出答案即可。

AC代码

#include<bits/stdc++.h>#define ll long long#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)#define fil(a,b) memset((a),(b),sizeof(a))#define cl(a) fil(a,0)#define PI 3.1415927#define inf 0x3f3f3f3fusing namespace std;int main(){    int t;    double n;    double from,to,mid;    cin>>t;    while(t--)    {        from=0;        to=100;        cin>>n;        if(!(n>=6&&n<=807020306)) cout<<"No solution!"<<endl;        else if(n==6) cout<<"0.0000"<<endl;        else if(n==807020306) cout<<"100.0000"<<endl;        else        {            while(to-from>1e-6)            {                mid=from+(to-from)/2;                if(8*pow(mid,4)+7*pow(mid,3)+2*pow(mid,2)+3*mid+6>n) to=mid;                else from=mid;            }            printf("%.4lf\n",to);        }    }    return 0;}