Strange fuction
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9155 Accepted Submission(s): 6236
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.
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
2 100 200
Sample Output
-74.4291 -178.8534
思路：求导得42.0 * pow(x,6) + 48*pow(x,5) + 21*pow(x,2) + 10*x - y ,那么此式子 = 0 时原函数最小。二分可解。得出的不是答案，还需要带入原函数得出结果。
Code：
#include <bits/stdc++.h>
#define LL long long
#define INF 1e-6
using namespace std;
double y;
double f ( double x ){
double ans = 42.0 * pow(x,6) + 48*pow(x,5) + 21*pow(x,2) + 10*x;
return ans ;
}
double res( double x , double y ){
double ans = 6 * pow(x,7) + 8 *pow(x,6) + 7*pow(x,3) + 5 *x*x - y*x;
return ans ;
}
int main(){
int n;
cin >> n ;
while( n-- ){
cin >> y ;
double l = 0.0 , r = 100.0;
while( r - l > 1e-7 ){
double mid = ( l + r ) / 2.0;
double re = f(mid);
if( re > y ){
r = mid - 1e-7;
}else{
l = mid + 1e-7 ;
}
}
printf("%.4lf\n", res( ((l+r)/2.0),y) );
}
return 0 ;
}