2199 Can you solve this equation?
2015-09-16 21:23
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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
Sample Output
Author
Redow
二分查找题~
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!
Author
Redow
二分查找题~
<span style="font-size:10px;">//HDOJ_2199 //20150916 #include<iostream> #include<cmath> #include<iomanip> using namespace std; double f(double x) { return 8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*pow(x,1)+6; } int main() { int T; double Y,l,r,m; cin>>T; while(T--) { cin>>Y; if(f(0)<=Y&&f(100)>=Y) { l=0; r=100; while(l-r<-1e-6) { m=(l+r)/2; double ans=f(m); if(ans>Y) r=m-(1e-7); else l=m+(1e-7); } cout<<fixed<<setprecision(4)<<(l+r)/2<<endl; } else cout<<"No solution!"<<endl; } }</span>
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