Strange fuction
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4803 Accepted Submission(s): 3428
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
Author
Redow
题意:给一式子,还有变量范,0-100,让求最小值,。导数等于0时,二分查找导数等于0的时候的x~我怎么把二分这么重要的查找方式给忘了呢,当时还在想这x是实数怎么找啊~蠢到猪
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 5 using namespace std; 6 7 double y; 8 9 double slove(double x) 10 { 11 return 42.0*x*x*x*x*x*x+48.0*x*x*x*x*x+21.0*x*x+10.0*x-y; 12 } 13 double get(double x) 14 { 15 return 6.0*x*x*x*x*x*x*x+8.0*x*x*x*x*x*x+7.0*x*x*x+5.0*x*x-y*x; 16 } 17 18 int main() 19 { 20 int t; 21 scanf("%d", &t); 22 while(t--) 23 { 24 scanf("%lf", &y); 25 double l = 0.0, r = 100.0; 26 while(r-l >= 1e-6) // double 比较,不能写成 r > l 27 { 28 double mid = (l+r)/2; 29 if(slove(mid) > 0) 30 r = mid; 31 else 32 l = mid; 33 } 34 printf("%.4f ", get(l)); 35 } 36 return 0; 37 }
蠢成猪肿么办~