hdu2899——Strange fuction

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3344    Accepted Submission(s): 2446

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

 

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对这个函数连续求几次导可以发现，对于此函数的导函数，如果y <= 0,那么函数在[0, 100]内单调递增，所以最小值是0
如果y>0,观察其导函数，发现导函数单调递增，所以我们只要找到导函数的零点（即原函数的极小值点），此时可以用二分解决
找到极值点以后，如果它在[0, 100]，就把它代入函数，否则说明函数在[0, 100]内单调递减，把100代入函数即可

#include <map>  #include <set>  #include <list>  #include <stack>  #include <queue>  #include <vector>  #include <cmath>  #include <cstdio>  #include <cstring>  #include <iostream>  #include <algorithm>using namespace std;int main(){int t;scanf("%d\n", &t);while (t--){double y;scanf("%lf", &y);if (y <= 0){printf("0.0000\n");continue;}double l = 0, r = 100, mid;while(r - l > 1e-6){mid = (l + r) / 2;double dx = 42.0 * pow(mid, 6) + 48.0 *  pow(mid, 5) + 21.0 *  pow(mid, 2) + 10.0 * mid;if (abs(dx - y) < 1e-6){break;}else if(dx - y > 1e-6){r = mid;}else{l = mid;}}if (mid - 100.0 > 1e-6){mid = 100.0;}double ans = 6 *  pow(mid, 7) + 8 *  pow(mid, 6) + 7 * pow(mid, 3) + 5 *  pow(mid, 2) - y * mid; printf("%.4f\n", ans);}return 0;}