HDU 2199 Can you solve this equation?
 
 
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.

InputThe 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);OutputFor 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! 这种题一般都很注重精度的,可以拿两个端点来做验证,6的时候应该为0.0000,807020306的时候应该为100.0000
其中while(ri-le<=0.000000001)这里的0尽量多一点,而判断的时候的0的个数要看情况,有些当满足
ri-le<=0.000000001时,说不定判断还不满足。
#include <iostream>
#include <stack>
#include <string.h>
#include <stdio.h>
#include<queue>
#include<algorithm>
#define ll long long
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
double y;
cin>>y;
double le,ri,mid;
le=;ri=;
bool f=;
while(ri-le>=0.00000000000001)
{
mid=(le+ri)/;
//cout<<8*mid*mid*mid*mid+7*mid*mid*mid+2*mid*mid+3*mid+6-y<<endl;
if(*mid*mid*mid*mid+*mid*mid*mid+*mid*mid+*mid+-y<0.00001&&*mid*mid*mid*mid+*mid*mid*mid+*mid*mid+*mid+-y>=)
{
f=;
break;
}
else if(*mid*mid*mid*mid+*mid*mid*mid+*mid*mid+*mid+-y<)
{
le=mid;
}
else
ri=mid;
}
if(f)
printf("%.4lf\n",mid);
else
printf("No solution!\n");
}
return ;
}
 

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