Problem F

Solve It

Input: standard input

Output: standard output

Time Limit: 1 second

Memory Limit: 32 MB

Solve the equation:
        p*e^-x+ q*sin(x) + r*cos(x) + s*tan(x) + t*x² + u = 0
        where 0 <= x <= 1.

Input

Input consists of multiple test cases and terminated by an EOF. Each test case consists of 6 integers in a single line: p, q, r, s, t and u (where 0 <= p,r<= 20 and -20 <= q,s,t <= 0). There will be maximum 2100 lines in the input file.

Output

For each set of input, there should be a line containing the value of x, correct upto 4 decimal places, or the string "No solution", whichever is applicable.

Sample Input

0 0 0 0 -2 1

1 0 0 0 -1 2

1 -1 1 -1 -1 1

Sample Output

0.7071

No solution

0.7554

题意：找到一个数X满足 p*e-x+ q*sin(x) + r*cos(x) + s*tan(x) + t*x2 + u = 0

代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

using namespace std;

double p,q,r,s,t,u;

double findx(double x)
{
    return p*exp(-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*x*x+u;
}

int main()
{
    while(scanf("%lf%lf%lf%lf%lf%lf",&p,&q,&r,&s,&t,&u)!=EOF)
    {
        if(findx(0)<0 || findx(1)>0)
        {
            printf("No solution\n");
        }
        else
        {
            double x1 = 0;
            double x2 = 1;
            while(fabs(x1-x2)>=1e-10)
            {
                double x = (x1+x2)/2;
                if(findx(x)>0)
                {
                    x1 = x;
                }
                else
                {
                    x2 = x;
                }
            }
            printf("%.4lf\n",x1);
        }
    }
    return 0;
}

Mustaq Ahmed