UVa 10341 - Solve It

链接：

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

原题：

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

分析与总结：

非线性方程求根问题， LRJ《算法入门经典》p150有类似的问题。  要求的跟是0~1之间， 而且这个方程是单调递减的，所以可以用二分来求根。

/*
 * UVa: 10341 - Solve It
 * Time: 0.024s
 * Result: Accept
 * Author: D_Double
 *
 */
#include<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;
}

——  生命的意义，在于赋予它意义。
   
    
   
   
   

     
    
    
              
    
   
   
   
    
   
   
   

     
    
    
         原创 http://blog.csdn.net/shuangde800 ， By   D_Double  (转载请标明)