二分查找求函数的区间最小值&&http://acm.hdu.edu.cn/showproblem.php?pid=2899
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Strange fuction
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 792 Accepted Submission(s): 607
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
2100200
Sample Output
-74.4291-178.8534AC代码:#include<iostream>#include<cmath>#include<cstdio>#define exp 1e-7using namespace std;double f(double x){ return 42*pow(x,6.0)+48*pow(x,5.0)+21*pow(x,2.0)+10*x;}double f1(double x,int a){return 6*pow(x,7.0)+8*pow(x,6.0)+7*pow(x,3.0)+5*pow(x,2.0)-a*x;}int main(){int T;scanf("%d",&T);while(T--){int a;scanf("%d",&a); double l=0.0,r=100.0,mid;while(r-l>exp){ mid=(l+r)/2; if(f(mid)>a) r=mid-exp; else l=mid+exp;}printf("%.4lf\n",f1(mid,a));}return 0;}
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