uva 10341 Solve It

二分法，注意端点精度0.00001是过不了的……1e-9差不多吧……

#include<stdio.h>
#include<math.h>
double a1,a2,p,q,r,s,t,u;

double f(double x)
{
    double ans=p*exp(-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*x*x+u;
    return ans;
}
int main()
{
    double ans;
    while(~scanf("%lf%lf%lf%lf%lf%lf",&p,&q,&r,&s,&t,&u))
    {
        a1=0;a2=1;
        if(f(0)<0&&f(1)<0) printf("No solution
");
        else if(f(0)>0&&f(1)>0) printf("No solution
");
        else
        {
            while(a2-a1>=0.000000001)
            {
                ans=(a2-a1)/2;
                if(f(ans+a1)>0)
                {
                    if(f(a1)<0) a2=ans+a1;
                    else a1=ans+a1;
                }
                else
                {
                    if(f(a1)<0) a1=ans+a1;
                    else a2=ans+a1;
                }
            }
            printf("%.4f
",a1);
        }
    }
    return 0;
}

简化版

#include<stdio.h>
#include<math.h>
double p,q,r,s,t,u;

double f(double x)
{
    double ans=p*exp(-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*x*x+u;
    return ans;
}
int main()
{
    double a1,a2,ans,k;
    while(~scanf("%lf%lf%lf%lf%lf%lf",&p,&q,&r,&s,&t,&u))
    {
        a1=0;a2=1;
        if(f(0)*f(1)>0) printf("No solution
");
        else
        {
            while(a2-a1>0.000000001)
            {
                ans=a1+(a2-a1)/2;
                k=f(ans);
                if(k<0) a2=ans;
                else a1=ans;
            }
            printf("%.4lf
",a1);
        }
    }
    return 0;
}

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