hdu 2199 Can you solve this equation? 二分
2018-02-27 18:37
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1.精度问题 由于是double类型,r=mid 而不是r=mid-1
2.如果首位两端(f(0)和f(100))同号,证明解不在[1,100]区间内 这是我之所以TE的原因,没有预先判断
3.若在这个区间内,则一定可要求出解 所以binarysearch 返回m
#include <time.h>
#include <iomanip>
#include <cstdlib>
#include <stdio.h>
#define eps 1e-6
using namespace std;
double y;
double f(double x)
{
return 8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6-y;
}
double binaryserch(double l, double r)
{
double m;
while((r-l)>=eps)
{
m=(l+r)/2;
if(f(m)>0) r=m;
else l=m;
}
return m;
}
int main()
{
int num;
cin>>num;
while(num--)
{
cin>>y;
if(f(0)*f(100)>0) cout<<"No solution!"<<endl;
else printf("%.4lf\n",binaryserch(0,100));
}
return 0;
}
2.如果首位两端(f(0)和f(100))同号,证明解不在[1,100]区间内 这是我之所以TE的原因,没有预先判断
3.若在这个区间内,则一定可要求出解 所以binarysearch 返回m
#include <time.h>
#include <iomanip>
#include <cstdlib>
#include <stdio.h>
#define eps 1e-6
using namespace std;
double y;
double f(double x)
{
return 8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6-y;
}
double binaryserch(double l, double r)
{
double m;
while((r-l)>=eps)
{
m=(l+r)/2;
if(f(m)>0) r=m;
else l=m;
}
return m;
}
int main()
{
int num;
cin>>num;
while(num--)
{
cin>>y;
if(f(0)*f(100)>0) cout<<"No solution!"<<endl;
else printf("%.4lf\n",binaryserch(0,100));
}
return 0;
}
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