hdu 2899 Strange fuction 【二分+數學函式求導】
阿新 • • 發佈:2019-01-29
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10131 Accepted Submission(s): 6835
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
2
100
200
Sample output
-74.4291
-178.8534
分析:這是一道數學題,也是第一道整理的二分題,開個頭!說白了,這道題就是讓你用數學知識求函式最小值。二分對精度控制一下即可。程式碼:
#include<iostream> #include<algorithm> #include<string.h> using namespace std; double solve(double x,double y) { return 6*pow(x,7) + 8*pow(x,6) + 7*pow(x,3) + 5*pow(x,2) - y*x; } double w(double x,double y) { return 42*pow(x,6) + 48*pow(x,5) + 21*pow(x,2) + 10*x - y; } int main() { ios::sync_with_stdio(false); int t; double l,r,mid; //第一次定義成了int,輸出結果都不顯示 double y; cin>>t; while(t--) { cin>>y; l = 0; r = 100; mid = (l+r)/2; while(fabs(w(mid,y)) > 0.000001) { if(w(mid,y) >= 0) //這種情況說明一階倒數等於0的根在(l,mid)之間 { r = mid; mid = (l+r)/2; } else //這種情況說明一階倒數等於0的根在(mid,r)之間 { l = mid; mid = (l+r)/2; } } printf("%.4f\n",solve(mid,y)); //注意在G++中提交用%f } return 0; }