UVa 10341： Solve It

UVa 10341： Solve It2014-12-02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
"）; continue;}while（right-left > EPS）{mid = （left+right）/2;if（fomula（mid）*fomula（left） > 0） left=mid;else right=mid;}printf（"%.4f
", mid）;} return 0;}

作者：csdn博客 shuangde800