单调函数零点可用二分法求
基本模板
double f(double a){} //单调函数
....
while(right-left>1e-10){
mid=(right+left)/2;
if(f(mid)>0) left=mid; //根据是递增还是递减判断
else right=mid;
}
对于凹函数或凸函数的极值可用三分法求,当然也可先求导,再二分求极值点
基本模板
double f(double a){} //函数
....
while(right-left>1e-10){
mid=(right+left)/2;
mmid=(mid+right)/2;
if(f(mid)>f(mmid)) right=mid; //根据是凹还是凸判断
else left=mid;
}
Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,can you find its solution between 0 and 100;
Now please try your lucky.
Now please try your lucky.
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 a real number Y (fabs(Y) <= 1e10);Output For each test case, you should just output one real number(accurate up to 4 decimal places),which is the solution of the equation,or “No solution!”,if there is no solution for the equation between 0 and 100.Sample Input
2 100 -4Sample Output
1.6152 No solution!
思路分析:当函数在区间与X轴没交点时,无解
#include<iostream>
#include<cmath>
#include<stdio.h>
using namespace std;
double c;
double f(double x){
double y;
y=8*pow(x,4)+7*pow(x,3)+2*x*x+3*x+c;
return y;
}
int main(){
double y,a,b,mid;
int t;
cin>>t;
while(t--){
cin>>y;
c=6-y;
if(f(100)<=0||f(0)>=0) cout<<"No solution!"<<endl;
else{
a=0;
b=100;
mid=(a+b)/2;
while(abs(b-a)>0.0000001){
if(f(mid)>0) b=mid;
else a=mid;
mid=(a+b)/2;
}
printf("%.4f\n",mid);
}
}
return 0;
}
题目2:hdoj2899
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.
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 200Sample Output
-74.4291 -178.8534
#include<iostream>
#include<cmath>
#include<stdio.h>
using namespace std;
double y;
double f1(double x){ //导函数
double ans;
ans=42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y;
return ans;
}
double f2(double x){ //原函数
double ans;
ans=6*pow(x,7)+8*pow(x,6)+7*x*x*x+5*x*x-y*x;
return ans;
}
int main(){
double a,b,mid;
int t;
cin>>t;
while(t--){
cin>>y;
if(f1(0)>=0) printf("%.4f\n",f2(0));
if(f1(100)<=0) printf("%.4f\n",f2(100));
else{
a=0;
b=100;
mid=(a+b)/2;
while(abs(b-a)>0.00000001){
if(f1(mid)>0) b=mid;
else a=mid;
mid=(a+b)/2;
}
printf("%.4f\n",f2(a));
}
}
return 0;
}