HDU2899 Strange fuction 【二分】
阿新 • • 發佈:2017-06-07
lines math.h only include return cti ack onos std
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
Sample Output
Strange fuction
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2982 Accepted Submission(s): 2202Problem 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
#include <stdio.h> #include <math.h> double y, ans; double f(double x){ return 6 * pow(x, 7) + 8 * pow(x, 6) + 7 * pow(x, 3) + 5 * x * x - y * x; } double ff(double x){ //求導 return 42 * pow(x, 6) + 48 * pow(x, 5) + 21 * x * x + 10 * x - y; } int main(){ int n; scanf("%d", &n); while(n--){ scanf("%lf", &y); if(ff(0) >= 0) ans = f(0); else if(ff(100) <= 0) ans = f(100); else{ double left = 0, right = 100, mid; while(left + 1e-8 < right){ mid = (left + right) / 2; if(ff(mid) > 0) right = mid; else if(ff(mid) < 0) left = mid; else break; } ans = f(mid); } printf("%.4lf\n", ans); } return 0; }
HDU2899 Strange fuction 【二分】