【精度问题】【HDU2899】Strange fuction

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3809    Accepted Submission(s): 2760

Problem Description

Now, here is a fuction:
  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
Can you find the minimum value when x is between 0 and 100.

 

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 only one real numbers Y.(0 < Y <1e10)

 

Output

Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.

 

Sample Input

2100200

 

Sample Output

-74.4291-178.8534

 

Author

Redow

 

很水的题 但是WA了几次

虽说最后要求1e-4的精度 

但是因为求最小值 所以注意所求的x值精度 至少要1e-6 所以请注意

#include <cstdio>  #include <cstdlib>  #include <cmath>  #include <cstring>  #include <ctime>  #include <algorithm>  #include <iostream>#include <sstream>#include <string>#define oo 0x13131313   using namespace std;double y;void init(){freopen("a.in","r",stdin);freopen("a.out","w",stdout);}double cal1(double x){return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;}double  cal2(double x){return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y;}void solve(){double L=0,R=100,ans=1e10; double temp1=cal1(L),temp2=cal1(R);if(temp1<ans) ans=temp1;if(temp2<ans) ans=temp2;double ANS;while(R-L>1e-6){double m=(R+L)/2;if(cal2(m)>0) R=m;else if(cal2(m)<0) L=m;else { ANS=m;break; }}ANS=R;if(cal1(ANS)<ans) ans=cal1(ANS);printf("%.4lf\n",ans);}int main(){//init();int T;cin>>T;while(T--){cin>>y;solve();}return 0;}