UVa 10341:solve It

Link:

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=113&page=show_ problem&problem=1282

Original title:

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-a single line:p, Q, R, S, T and U (where 0 <= p,r <=-and-20 <= <= 0). There is maximum 2100 lines in the input file.

Output

For each set of input, there should is 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

Analysis and Summary:

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The problem of solving the root of nonlinear equation, lrj the classic of algorithm p150 has similar problems. The requirement is between the 0~1, and the equation is monotonically decreasing, so you can use the two points to find the root.

 
 * * Uva:10341-solve It 
 * time:0.024s 
 * author:d_double * * * * * * * * * * * * * */
#include &l t;iostream>  
#include <cstdio>  
#include <cmath>  
#define EPS (10e-8)  
using namespace STD;  
      
Double p,q,r,s,t,u;  
      
Inline double Fomula (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) {  
          
        double Left=0, Right=1, mid;  
        BOOL Flag=false;  
        if (Fomula (left) *fomula (right) > 0) {  
            printf ("No solution\n");   
            Continue;  
        }  
        while (Right-left > EPS) {  
            mid = (left+right)/2;  
            if (Fomula (mid) *fomula (left) > 0) left=mid;  
            else Right=mid;  
        }  
          
        printf ("%.4f\n", mid);  
    }   
    return 0;  
}

Author: csdn Blog shuangde800