ACM : HDU 2899 Strange fuction 解题报告 -二分、三分

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Strange fuction   Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5933    Accepted Submission(s): 4194     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      Author Redow      Recommend lcy  <br><br>//三分二分都可以，导数单调，求导后二分值 AC代码

　　

#include"iostream"
#include"algorithm"
#include"cstdio"
#include"cstring"
#include"cmath"
#define max(a,b) a>b?a:b
#define min(a,b) a<b?a:b
#define MX 10000 + 50
using namespace std;
double y;

double f(double x) {              //求导
    return 42*pow(x,6)+48*pow(x,5)+21*x*x+10*x-y; //  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
}

double F(double x) {
    return 6 * pow(x,7)+8*pow(x,6)+7*x*x*x+5*x*x-y*x;//  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
}

int main() {
    int T;
    while(cin>>T) {
        for(int qq=0; qq<T; qq++) {
            cin>>y;
            if(f(0)>0) {
                printf("%.4f\n",F(0));
            } else if(f(100)<0) {
                printf("%.4f\n",F(100));
            } else {
                double l=0;                //     0     l                r    100
                double r=100;           //   -------------------------
                double mid;
                while(r-l>1e-8) {
                    mid=(r+l)/2.0;
                    if(f(mid)>0) {
                        r=mid;
                    } else {
                        l=mid;
                    }
                }
                printf("%.4f\n",F(mid));
            }
        }
    }
    return 0;
}