HDU-2899

2019-11-09 20:10:23

字體：大中小

供稿：網(wǎng)友

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.InputThe 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)OutputJust the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.Sample Input

2100200Sample Output-74.4291-178.8534
這道題在做的時(shí)候，最開(kāi)始覺(jué)得這個(gè)函數(shù)應(yīng)該是用三分法解決，但是用了之后，會(huì)發(fā)現(xiàn)怎么都ac不了，都是
wa，然后在網(wǎng)上查完資料之后，發(fā)現(xiàn)這道題就應(yīng)該是先求導(dǎo)，之后再根據(jù)導(dǎo)數(shù)的正負(fù)，來(lái)運(yùn)用二分法解決問(wèn)
題，因?yàn)槲覀円蟮氖且粋€(gè)最值而跟極值的概念還是有不同的，之后將過(guò)程實(shí)現(xiàn)，就可以ac了，這道題感覺(jué)
就是可以運(yùn)用數(shù)學(xué)知識(shí)解決問(wèn)題
#include<iostream>#include<cmath>#include<cstdio>using namespace std;double cal(double x,double y){    //6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)    return(6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x);}double cal_d(double x,double y){    return(42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y);}int main(){    int times;    cin>>times;    while(times--)    {        long long y_get;        cin>>y_get;        double low = 0,high = 100;        double mid;        while(low<high-0.000001)        {            mid = (low+high)/2;            if(cal_d(mid,y_get)>0) high = mid;            else low = mid;        }        PRintf("%.4f/n",cal(mid,y_get));    }}