Strange fuctionTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4527Accepted 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
2100200
Sample Output
-74.4291-178.8534
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
//三分求极值#include <cmath>#include <cstdio>#include <string>#include <cstring>#include <iostream>#include <algorithm>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\n",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 <cmath>#include <cstdio>#include <string>#include <cstring>#include <iostream>#include <algorithm>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\n",F(mid));}int main() { // freopen("input.in","r",stdin);for(scanf("%d",&T);T–;){scanf("%lf",&Y);Search();}return 0;}
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慢慢学会了长大。流转的时光，
