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Strange fuction
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4527 Accepted Submission(s): 3251
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
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这是一个开口向上的函数，我们只要求这个函数的极（最）小值即可，可以三分直接求，当然也可以先把函数求导，然后二分求解，水题...附上两种解法
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//三分求极值
#include
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const double INF = 0x7fffffff; const double eps = 1e-10; int T; double Y,X; double Calc(double x) { return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-Y*x; } void Search() { double l,r,mid,Rmid,MVal,RMVal; l = 0;r = 100; while(r-l>=eps) { mid = (l+r)/2.0;Rmid = (mid+r)/2.0; MVal = Calc(mid);RMVal = Calc(Rmid); if(MVal < RMVal) { r = Rmid; } else if(MVal > RMVal) { l = mid; } else { r = mid; l = Rmid; } } printf(%.4lf ,Calc(mid)); } int main() { //freopen(input.in,r,stdin); for(scanf(%d,&T);T--;) { scanf(%lf,&Y); Search(); } return 0; } //二分求极值 #include
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const double INF = 0x7fffffff; const double eps = 1e-10; int T; double Y,X; double Calc(double x) { return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*pow(x,1)-Y; } double F(double x) { return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-Y*x; } void Search(){ double l,r,mid,MVal; l = 0;r = 100; while(r-l>=eps) { mid = (l+r)/2.0; MVal = Calc(mid); if(MVal > 0) { r = mid; } else if(MVal < 0) { l = mid; } else break; } printf(%.4lf ,F(mid)); } int main() { // freopen(input.in,r,stdin); for(scanf(%d,&T);T--;) { scanf(%lf,&Y); Search(); } return 0; }
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